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Polynomials, Factor and Remainder Theorems

22 questions

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Question 1Question →

When the polynomial P(x)=3x3−kx2+4x−7P(x) = 3x^3 - kx^2 + 4x - 7 is divided by x−2x - 2, the remainder is 99. What is the value of kk?

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Answer: 4

Answer

The value of kk is 44.
According to the Remainder Theorem, dividing P(x)P(x) by x−2x - 2 leaves a remainder of P(2)P(2). Evaluating P(2)=3(2)3−k(2)2+4(2)−7P(2) = 3(2)^3 - k(2)^2 + 4(2) - 7 gives 25−4k25 - 4k. Setting 25−4k=925 - 4k = 9 and solving yields k=4k = 4.

Step-by-Step Solution

1
Apply the Remainder Theorem
The remainder when P(x)P(x) is divided by x−2x - 2 is P(2)P(2).
By the Remainder Theorem, dividing a polynomial P(x)P(x) by x−ax - a leaves a remainder equal to P(a)P(a).
2
Substitute x=2x = 2 into P(x)P(x)
P(2)=3(2)3−k(2)2+4(2)−7=25−4kP(2) = 3(2)^3 - k(2)^2 + 4(2) - 7 = 25 - 4k
Evaluating the polynomial at x=2x = 2 expresses the remainder in terms of kk.
3
Equate P(2)P(2) to the given remainder and solve for kk
25−4k=9  ⟹  4k=16  ⟹  k=425 - 4k = 9 \implies 4k = 16 \implies k = 4
Setting the calculated expression equal to 99 forms a linear equation that yields k=4k = 4.

Key Concept

Polynomial Remainder Theorem
Estimated Time:1m 30s
Question 2Question →

What is the remainder when the polynomial P(x)=2x3−5x2+7x−3P(x) = 2x^3 - 5x^2 + 7x - 3 is divided by (2x−1)(2x - 1)?

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Answer: −12-\frac{1}{2}

Answer

The remainder is −12-\frac{1}{2}.
By the Remainder Theorem, when P(x)P(x) is divided by (ax−b)(ax - b), the remainder is P(ba)P\left(\frac{b}{a}\right). Setting 2x−1=02x - 1 = 0 yields x=12x = \frac{1}{2}. Evaluating P(12)=2(18)−5(14)+7(12)−3=14−54+72−3=−12P\left(\frac{1}{2}\right) = 2\left(\frac{1}{8}\right) - 5\left(\frac{1}{4}\right) + 7\left(\frac{1}{2}\right) - 3 = \frac{1}{4} - \frac{5}{4} + \frac{7}{2} - 3 = -\frac{1}{2}.

Step-by-Step Solution

1
Find the root of the linear divisor
2x−1=0  ⟹  x=122x - 1 = 0 \implies x = \frac{1}{2}
According to the Remainder Theorem, dividing P(x)P(x) by a linear divisor (ax−b)(ax - b) yields a remainder of P(ba)P\left(\frac{b}{a}\right).
2
Substitute x=12x = \frac{1}{2} into P(x)=2x3−5x2+7x−3P(x) = 2x^3 - 5x^2 + 7x - 3
P(12)=2(12)3−5(12)2+7(12)−3P\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - 5\left(\frac{1}{2}\right)^2 + 7\left(\frac{1}{2}\right) - 3
Evaluating the polynomial at the root of the divisor determines the remainder.
3
Calculate the arithmetic value
P(12)=2(18)−5(14)+72−3=14−54+72−3=−1+72−3=−4+3.5=−0.5=−12P\left(\frac{1}{2}\right) = 2\left(\frac{1}{8}\right) - 5\left(\frac{1}{4}\right) + \frac{7}{2} - 3 = \frac{1}{4} - \frac{5}{4} + \frac{7}{2} - 3 = -1 + \frac{7}{2} - 3 = -4 + 3.5 = -0.5 = -\frac{1}{2}
Simplifying the fractional terms gives the final remainder value.

Key Concept

Remainder Theorem for Linear Divisors (ax−b)(ax - b)
Question 3Question →

What is the remainder when the polynomial P(x)=x3+3x2−2x+4P(x) = x^3 + 3x^2 - 2x + 4 is divided by x−1x - 1?

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Answer: 66

Answer

The remainder when P(x)P(x) is divided by x−1x - 1 is 66.
According to the Remainder Theorem, dividing a polynomial P(x)P(x) by a linear divisor x−ax - a leaves a remainder equal to P(a)P(a). For the divisor x−1x - 1, setting x−1=0x - 1 = 0 yields x=1x = 1. Substituting x=1x = 1 into P(x)=x3+3x2−2x+4P(x) = x^3 + 3x^2 - 2x + 4 gives 1+3−2+4=61 + 3 - 2 + 4 = 6. Therefore, the value 66 is the correct remainder.

Step-by-Step Solution

1
Apply the Remainder Theorem
To find the remainder when P(x)P(x) is divided by x−ax - a, set x−1=0x - 1 = 0, giving x=1x = 1. The remainder is equal to P(1)P(1).
By the Remainder Theorem, dividing a polynomial P(x)P(x) by (x−a)(x - a) yields a remainder of P(a)P(a).
2
Substitute x=1x = 1 into P(x)=x3+3x2−2x+4P(x) = x^3 + 3x^2 - 2x + 4
P(1)=(1)3+3(1)2−2(1)+4=1+3−2+4=6P(1) = (1)^3 + 3(1)^2 - 2(1) + 4 = 1 + 3 - 2 + 4 = 6.
Direct evaluation of the expression at x=1x = 1 yields the numerical value of the remainder.

Key Concept

The Remainder Theorem states that when a polynomial P(x)P(x) is divided by a linear factor (x−a)(x - a), the remainder is P(a)P(a).
Question 4Question →

The polynomial P(x)=2x3+px2+qx−6P(x) = 2x^3 + px^2 + qx - 6 has (x−2)(x - 2) as a factor. When P(x)P(x) is divided by (x+1)(x + 1), the remainder is −12-12. What is the value of p+qp + q?

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Answer: −2-2

Answer

The value of p+qp + q is −2-2.
Using the Factor Theorem, P(2)=0P(2) = 0 yields 2p+q=−52p + q = -5. Using the Remainder Theorem, P(−1)=−12P(-1) = -12 yields p−q=−4p - q = -4. Solving these two equations simultaneously gives p=−3p = -3 and q=1q = 1. Therefore, p+q=−3+1=−2p + q = -3 + 1 = -2.

Step-by-Step Solution

1
Apply the Factor Theorem for divisor (x−2)(x - 2)
2p+q=−52p + q = -5
Since (x−2)(x - 2) is a factor of P(x)P(x), P(2)=0P(2) = 0. Substituting x=2x = 2 gives 2(2)3+p(2)2+q(2)−6=0  ⟹  16+4p+2q−6=0  ⟹  4p+2q=−10  ⟹  2p+q=−52(2)^3 + p(2)^2 + q(2) - 6 = 0 \implies 16 + 4p + 2q - 6 = 0 \implies 4p + 2q = -10 \implies 2p + q = -5.
2
Apply the Remainder Theorem for divisor (x+1)(x + 1)
p−q=−4p - q = -4
Dividing P(x)P(x) by (x+1)(x + 1) leaves a remainder of −12-12, so P(−1)=−12P(-1) = -12. Substituting x=−1x = -1 gives 2(−1)3+p(−1)2+q(−1)−6=−12  ⟹  −2+p−q−6=−12  ⟹  p−q=−42(-1)^3 + p(-1)^2 + q(-1) - 6 = -12 \implies -2 + p - q - 6 = -12 \implies p - q = -4.
3
Solve the simultaneous linear equations for pp and qq
p=−3p = -3 and q=1q = 1
Adding the two equations (2p+q)+(p−q)=−5+(−4)(2p + q) + (p - q) = -5 + (-4) yields 3p=−9  ⟹  p=−33p = -9 \implies p = -3. Substituting p=−3p = -3 into p−q=−4p - q = -4 gives −3−q=−4  ⟹  q=1-3 - q = -4 \implies q = 1.
4
Calculate the required expression p+qp + q
p+q=−2p + q = -2
Summing the calculated constants: p+q=−3+1=−2p + q = -3 + 1 = -2.

Key Concept

Factor and Remainder Theorems for Polynomials
Estimated Time:2m 0s
Question 5Question →

When the polynomial P(x)=3x3+ax2+bx−10P(x) = 3x^3 + ax^2 + bx - 10 is divided by (x−2)(x - 2), the remainder is 1414, and when it is divided by (x+1)(x + 1), the remainder is −16-16. What is the value of a+ba + b?

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Answer: 1

Answer

The value of a+ba + b is 11.
According to the Remainder Theorem, dividing P(x)P(x) by (x−2)(x - 2) and (x+1)(x + 1) gives P(2)=14P(2) = 14 and P(−1)=−16P(-1) = -16 respectively. Expanding these expressions forms two linear equations: 2a+b=02a + b = 0 and a−b=−3a - b = -3. Solving these simultaneously gives a=−1a = -1 and b=2b = 2, so a+b=1a + b = 1.

Step-by-Step Solution

1
Apply the Remainder Theorem for the first divisor (x−2)(x - 2)
2a+b=02a + b = 0
By the Remainder Theorem, P(2)=14P(2) = 14. Substituting x=2x = 2 into P(x)P(x) yields 3(8)+4a+2b−10=143(8) + 4a + 2b - 10 = 14, which simplifies to 2a+b=02a + b = 0.
2
Apply the Remainder Theorem for the second divisor (x+1)(x + 1)
a−b=−3a - b = -3
By the Remainder Theorem, P(−1)=−16P(-1) = -16. Substituting x=−1x = -1 into P(x)P(x) yields 3(−1)+a−b−10=−163(-1) + a - b - 10 = -16, which simplifies to a−b=−3a - b = -3.
3
Solve the simultaneous equations for aa and bb
a=−1a = -1 and b=2b = 2
Adding 2a+b=02a + b = 0 and a−b=−3a - b = -3 yields 3a=−33a = -3, giving a=−1a = -1. Substituting a=−1a = -1 into 2a+b=02a + b = 0 gives b=2b = 2.
4
Calculate the value of a+ba + b
1
Adding the computed values yields a+b=−1+2=1a + b = -1 + 2 = 1.

Key Concept

Polynomial Remainder Theorem and Systems of Linear Equations
Estimated Time:2m 0s
Question 6Question →

When the polynomial P(x)=x3−ax2+bx−6P(x) = x^3 - ax^2 + bx - 6 is divided by (x−1)(x - 1), the remainder is −4-4. If (x−2)(x - 2) is a factor of P(x)P(x), what is the value of a+ba + b?

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Answer: 55

Answer

The value of a+ba + b is 55.
Using the Remainder Theorem with P(1)=−4P(1) = -4 gives the equation −a+b=1-a + b = 1. Using the Factor Theorem with P(2)=0P(2) = 0 gives 2a−b=12a - b = 1. Solving this simultaneous system gives a=2a = 2 and b=3b = 3, leading to a+b=5a + b = 5.

Step-by-Step Solution

1
Apply the Remainder Theorem for division by (x−1)(x - 1).
−a+b=1-a + b = 1
According to the Remainder Theorem, P(1)=−4P(1) = -4. Substituting x=1x = 1 into P(x)P(x) yields 13−a(1)2+b(1)−6=−4  ⟹  −a+b=11^3 - a(1)^2 + b(1) - 6 = -4 \implies -a + b = 1.
2
Apply the Factor Theorem for the factor (x−2)(x - 2).
2a−b=12a - b = 1
Since (x−2)(x - 2) is a factor, P(2)=0P(2) = 0. Substituting x=2x = 2 into P(x)P(x) yields 23−a(2)2+b(2)−6=0  ⟹  8−4a+2b−6=0  ⟹  2a−b=12^3 - a(2)^2 + b(2) - 6 = 0 \implies 8 - 4a + 2b - 6 = 0 \implies 2a - b = 1.
3
Solve the system of linear equations simultaneously.
a=2,b=3a = 2, b = 3
Adding the two equations (−a+b)+(2a−b)=1+1(-a + b) + (2a - b) = 1 + 1 yields a=2a = 2. Substituting a=2a = 2 back into −a+b=1-a + b = 1 gives b=3b = 3.
4
Calculate the required value a+ba + b.
55
a+b=2+3=5a + b = 2 + 3 = 5.

Key Concept

Polynomial Factor and Remainder Theorems
Estimated Time:1m 30s
Question 7Question →

When the polynomial P(x)=2x4+ax3+bx2−5x+6P(x) = 2x^4 + ax^3 + bx^2 - 5x + 6 is divided by (x−2)(x+1)(x - 2)(x + 1), the remainder is 6x+86x + 8. What is the value of a−ba - b?

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Answer: 11

Answer

11
According to the Remainder Theorem, dividing P(x)P(x) by (x−2)(x+1)(x - 2)(x + 1) leaves a remainder R(x)=6x+8R(x) = 6x + 8. Substituting x=−1x = -1 gives R(−1)=2R(-1) = 2 and P(−1)=2−a+b+5+6=13−a+bP(-1) = 2 - a + b + 5 + 6 = 13 - a + b. Equating 13−a+b=213 - a + b = 2 gives −a+b=−11-a + b = -11, which directly leads to a−b=11a - b = 11.

Step-by-Step Solution

1
Apply the Remainder Theorem for linear factors of the divisor (x−2)(x+1)(x - 2)(x + 1).
Since the divisor is (x−2)(x+1)(x - 2)(x + 1), the roots of the divisor are x=2x = 2 and x=−1x = -1. The remainder function is R(x)=6x+8R(x) = 6x + 8, so P(2)=R(2)P(2) = R(2) and P(−1)=R(−1)P(-1) = R(-1).
By the Polynomial Division Algorithm, P(x)=(x−2)(x+1)Q(x)+R(x)P(x) = (x - 2)(x + 1)Q(x) + R(x).
2
Evaluate R(x)R(x) and P(x)P(x) at x=−1x = -1.
R(−1)=6(−1)+8=2R(-1) = 6(-1) + 8 = 2.
P(−1)=2(−1)4+a(−1)3+b(−1)2−5(−1)+6=2−a+b+5+6=13−a+bP(-1) = 2(-1)^4 + a(-1)^3 + b(-1)^2 - 5(-1) + 6 = 2 - a + b + 5 + 6 = 13 - a + b.
Setting P(−1)=R(−1)P(-1) = R(-1) gives 13−a+b=213 - a + b = 2.
Substituting x=−1x = -1 eliminates the quotient term since (−1+1)=0(-1 + 1) = 0.
3
Rearrange the equation to solve for a−ba - b.
13−a+b=2  ⟹  −a+b=−11  ⟹  a−b=1113 - a + b = 2 \implies -a + b = -11 \implies a - b = 11.
Multiplying both sides of −a+b=−11-a + b = -11 by −1-1 gives a−b=11a - b = 11.

Key Concept

Remainder Theorem for Composite Linear Divisors
Question 8Question →

When the polynomial P(x)=2x3+3x2−px+qP(x) = 2x^3 + 3x^2 - px + q is divided by (x−1)(x - 1), the remainder is 33. Given that (x+2)(x + 2) is a factor of P(x)P(x), what is the value of p+qp + q?

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Answer: 2

Answer

The value of p+qp + q is 22.
According to the Remainder Theorem, P(1)=3P(1) = 3 gives 2(1)3+3(1)2−p(1)+q=32(1)^3 + 3(1)^2 - p(1) + q = 3, which simplifies to p−q=2p - q = 2. According to the Factor Theorem, (x+2)(x + 2) being a factor means P(−2)=0P(-2) = 0, giving 2(−2)3+3(−2)2−p(−2)+q=02(-2)^3 + 3(-2)^2 - p(-2) + q = 0, which simplifies to 2p+q=42p + q = 4. Solving these equations together gives p=2p = 2 and q=0q = 0. Summing them yields p+q=2p + q = 2.

Step-by-Step Solution

1
Apply the Remainder Theorem for divisor (x−1)(x - 1)
P(1)=2(1)3+3(1)2−p(1)+q=3  ⟹  5−p+q=3  ⟹  p−q=2P(1) = 2(1)^3 + 3(1)^2 - p(1) + q = 3 \implies 5 - p + q = 3 \implies p - q = 2
By the Remainder Theorem, dividing P(x)P(x) by (x−a)(x - a) leaves a remainder equal to P(a)P(a).
2
Apply the Factor Theorem for factor (x+2)(x + 2)
P(−2)=2(−2)3+3(−2)2−p(−2)+q=0  ⟹  −16+12+2p+q=0  ⟹  2p+q=4P(-2) = 2(-2)^3 + 3(-2)^2 - p(-2) + q = 0 \implies -16 + 12 + 2p + q = 0 \implies 2p + q = 4
By the Factor Theorem, if (x−a)(x - a) is a factor of P(x)P(x), then P(a)=0P(a) = 0. Here a=−2a = -2.
3
Solve the simultaneous linear equations for pp and qq
Adding p−q=2p - q = 2 and 2p+q=42p + q = 4 yields 3p=6  ⟹  p=23p = 6 \implies p = 2. Substituting p=2p = 2 into p−q=2p - q = 2 gives q=0q = 0.
Eliminating qq allows direct calculation of pp, followed by back-substitution for qq.
4
Calculate the target expression p+qp + q
p+q=2+0=2p + q = 2 + 0 = 2
Combine the values of pp and qq to obtain the required sum.

Key Concept

Factor and Remainder Theorems for Polynomials
Question 9Question →

When the polynomial P(x)=x3−2x2+ax+8P(x) = x^3 - 2x^2 + ax + 8 is divided by (x−3)(x - 3), the remainder is 1414. What is the value of the constant aa?

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Answer: -1

Answer

The value of the constant aa is −1-1.
According to the Remainder Theorem, dividing P(x)P(x) by (x−3)(x - 3) means the remainder is P(3)P(3). Evaluating P(3)=33−2(3)2+3a+8=27−18+3a+8=17+3aP(3) = 3^3 - 2(3)^2 + 3a + 8 = 27 - 18 + 3a + 8 = 17 + 3a. Setting this equal to the remainder 1414 gives 17+3a=1417 + 3a = 14, which simplifies to 3a=−33a = -3 and yields a=−1a = -1.

Step-by-Step Solution

1
Apply the Remainder Theorem
The remainder when P(x)P(x) is divided by (x−3)(x - 3) is equal to P(3)P(3).
The Remainder Theorem states that dividing a polynomial P(x)P(x) by (x−c)(x - c) yields a remainder equal to P(c)P(c).
2
Substitute x=3x = 3 into the polynomial and set equal to the given remainder
33−2(3)2+a(3)+8=143^3 - 2(3)^2 + a(3) + 8 = 14
Setting the value of P(3)P(3) equal to 1414 allows us to form a linear equation for the unknown constant aa.
3
Simplify numerical terms in the equation
27−18+3a+8=14  ⟹  17+3a=1427 - 18 + 3a + 8 = 14 \implies 17 + 3a = 14
Evaluate exponents and multiplication to isolate the term containing aa.
4
Solve for aa
3a=−3  ⟹  a=−13a = -3 \implies a = -1
Subtract 1717 from both sides and divide by 33.

Key Concept

Polynomial Remainder Theorem
Question 10Question →

If the polynomial P(x)=x4+ax3−7x2+bx+12P(x) = x^4 + ax^3 - 7x^2 + bx + 12 is completely divisible by x2−2x−3x^2 - 2x - 3, what is the value of a−ba - b?

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Answer: −10-10

Answer

The value of a−ba - b is −10-10.
Factoring x2−2x−3x^2 - 2x - 3 gives (x−3)(x+1)(x - 3)(x + 1). By the Factor Theorem, P(3)=0P(3) = 0 and P(−1)=0P(-1) = 0. Substituting these into P(x)P(x) produces the linear system 9a+b=−109a + b = -10 and a+b=6a + b = 6. Solving this system yields a=−2a = -2 and b=8b = 8. Subtracting gives a−b=−2−8=−10a - b = -2 - 8 = -10.

Step-by-Step Solution

1
Factor the quadratic divisor to find the roots.
x2−2x−3=(x−3)(x+1)x^2 - 2x - 3 = (x - 3)(x + 1), so the roots are x=3x = 3 and x=−1x = -1.
By the Factor Theorem, if a polynomial is divisible by a quadratic expression, P(x)P(x) must evaluate to zero at each root of the divisor.
2
Set up equations by evaluating P(3)=0P(3) = 0 and P(−1)=0P(-1) = 0.
For x=3x = 3: 34+a(3)3−7(3)2+b(3)+12=0  ⟹  81+27a−63+3b+12=0  ⟹  9a+b=−103^4 + a(3)^3 - 7(3)^2 + b(3) + 12 = 0 \implies 81 + 27a - 63 + 3b + 12 = 0 \implies 9a + b = -10.
For x=−1x = -1: (−1)4+a(−1)3−7(−1)2+b(−1)+12=0  ⟹  1−a−7−b+12=0  ⟹  a+b=6(-1)^4 + a(-1)^3 - 7(-1)^2 + b(-1) + 12 = 0 \implies 1 - a - 7 - b + 12 = 0 \implies a + b = 6.
Evaluating the polynomial at each root yields a system of two linear equations in variables aa and bb.
3
Solve the simultaneous equations for aa and bb.
Subtracting (a+b=6)(a + b = 6) from (9a+b=−10)(9a + b = -10) gives 8a=−16  ⟹  a=−28a = -16 \implies a = -2.
Substituting a=−2a = -2 into a+b=6a + b = 6 gives −2+b=6  ⟹  b=8-2 + b = 6 \implies b = 8.
Elimination isolates aa, allowing both aa and bb to be uniquely determined.
4
Calculate a−ba - b.
a−b=−2−8=−10a - b = -2 - 8 = -10.
This computes the required expression value.

Key Concept

Factor Theorem for quadratic divisors
Estimated Time:2m 0s
Question 11Question →

If (x+3)(x + 3) is a factor of the polynomial P(x)=x3+4x2+mx−6P(x) = x^3 + 4x^2 + mx - 6, what is the value of mm?

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Answer: 1

Answer

The value of mm is 1.
According to the Factor Theorem, (x+3)(x + 3) is a factor of P(x)P(x) if P(−3)=0P(-3) = 0. Substituting x=−3x = -3 into P(x)=x3+4x2+mx−6P(x) = x^3 + 4x^2 + mx - 6 yields (−3)3+4(−3)2+m(−3)−6=0(-3)^3 + 4(-3)^2 + m(-3) - 6 = 0, which simplifies to −27+36−3m−6=0-27 + 36 - 3m - 6 = 0, giving 3−3m=03 - 3m = 0 and thus m=1m = 1.

Step-by-Step Solution

1
Apply the Factor Theorem
P(−3)=0P(-3) = 0
By the Factor Theorem, for a linear divisor (x−a)(x - a) to be a factor of P(x)P(x), P(a)P(a) must equal zero. Here x+3=0  ⟹  x=−3x + 3 = 0 \implies x = -3.
2
Substitute x=−3x = -3 into P(x)=x3+4x2+mx−6P(x) = x^3 + 4x^2 + mx - 6
(-3)^3 + 4(-3)^2 + m(-3) - 6 = 0
Evaluating P(−3)P(-3) sets up an equation to find the unknown coefficient mm.
3
Simplify and solve for mm
-27 + 36 - 3m - 6 = 0 \implies 3 - 3m = 0 \implies m = 1
Combine the constant terms −27+36−6=3-27 + 36 - 6 = 3 and solve the linear equation in terms of mm.

Key Concept

Factor Theorem
Question 12Question →

When the cubic polynomial P(x)=2x3+px2+qx+6P(x) = 2x^3 + px^2 + qx + 6 is divided by x2−1x^2 - 1, the remainder is 3x+23x + 2. What is the remainder when P(x)P(x) is divided by 2x−32x - 3?

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Answer: 214\frac{21}{4}

Answer

The remainder when P(x)P(x) is divided by 2x−32x - 3 is 214\frac{21}{4}.
Using the Remainder Theorem on the quadratic divisor (x2−1)=(x−1)(x+1)(x^2 - 1) = (x - 1)(x + 1), we find P(1)=5P(1) = 5 and P(−1)=−1P(-1) = -1. Substituting these values into P(x)=2x3+px2+qx+6P(x) = 2x^3 + px^2 + qx + 6 produces the simultaneous linear equations p+q=−3p + q = -3 and p−q=−5p - q = -5. Solving gives p=−4p = -4 and q=1q = 1, leading to P(x)=2x3−4x2+x+6P(x) = 2x^3 - 4x^2 + x + 6. Dividing by (2x−3)(2x - 3) requires evaluating P(32)P\left(\frac{3}{2}\right), which equals 214\frac{21}{4}.

Step-by-Step Solution

1
Express the Division Algorithm for quadratic divisor x2−1=(x−1)(x+1)x^2 - 1 = (x - 1)(x + 1).
P(x)=(x2−1)Q(x)+(3x+2)P(x) = (x^2 - 1)Q(x) + (3x + 2)
By the Remainder Theorem, evaluating at roots of x2−1=0x^2 - 1 = 0 (x=1x = 1 and x=−1x = -1) yields the values of P(1)P(1) and P(−1)P(-1).
2
Calculate P(1)P(1) and P(−1)P(-1) from the remainder expression.
P(1)=3(1)+2=5P(1) = 3(1) + 2 = 5 and P(−1)=3(−1)+2=−1P(-1) = 3(-1) + 2 = -1
The quotient term (x2−1)Q(x)(x^2 - 1)Q(x) vanishes at x=1x = 1 and x=−1x = -1.
3
Substitute x=1x = 1 and x=−1x = -1 into P(x)=2x3+px2+qx+6P(x) = 2x^3 + px^2 + qx + 6 to build a system of linear equations.
P(1)=2+p+q+6=p+q+8=5  ⟹  p+q=−3P(1) = 2 + p + q + 6 = p + q + 8 = 5 \implies p + q = -3, and P(−1)=−2+p−q+6=p−q+4=−1  ⟹  p−q=−5P(-1) = -2 + p - q + 6 = p - q + 4 = -1 \implies p - q = -5
This sets up two simultaneous linear equations in terms of pp and qq.
4
Solve the system of equations for pp and qq.
Adding the equations gives 2p=−8  ⟹  p=−42p = -8 \implies p = -4, and substituting into p+q=−3p + q = -3 gives q=1q = 1. Thus, P(x)=2x3−4x2+x+6P(x) = 2x^3 - 4x^2 + x + 6.
Determining pp and qq gives the explicit formula for the polynomial.
5
Apply the Remainder Theorem to find the remainder when P(x)P(x) is divided by 2x−32x - 3.
Set 2x−3=0  ⟹  x=322x - 3 = 0 \implies x = \frac{3}{2}. Evaluate P(32)=2(32)3−4(32)2+32+6=2(278)−4(94)+32+6=274−9+64+6=334−3=214P\left(\frac{3}{2}\right) = 2\left(\frac{3}{2}\right)^3 - 4\left(\frac{3}{2}\right)^2 + \frac{3}{2} + 6 = 2\left(\frac{27}{8}\right) - 4\left(\frac{9}{4}\right) + \frac{3}{2} + 6 = \frac{27}{4} - 9 + \frac{6}{4} + 6 = \frac{33}{4} - 3 = \frac{21}{4}.
The remainder of a polynomial P(x)P(x) divided by (ax−b)(ax - b) is P(ba)P\left(\frac{b}{a}\right).

Key Concept

Polynomial Division Algorithm and Remainder Theorem for Linear and Quadratic Divisors
Question 13Question →

If (x−1)(x - 1) is a factor of the polynomial P(x)=x3+2x2−5x+kP(x) = x^3 + 2x^2 - 5x + k, what is the value of kk?

Show answer & explanation

Answer: 2

Answer

The value of kk is 22.
According to the Factor Theorem, a linear polynomial (x−a)(x - a) is a factor of P(x)P(x) if and only if P(a)=0P(a) = 0. For the factor (x−1)(x - 1), setting x=1x = 1 gives P(1)=(1)3+2(1)2−5(1)+k=0P(1) = (1)^3 + 2(1)^2 - 5(1) + k = 0. Simplifying yields 1+2−5+k=01 + 2 - 5 + k = 0, which simplifies further to −2+k=0-2 + k = 0, giving k=2k = 2.

Step-by-Step Solution

1
Apply the Factor Theorem
P(1)=0P(1) = 0
Since (x−1)(x - 1) is a factor, setting x=1x = 1 makes the polynomial equal to zero.
2
Substitute x=1x = 1 into P(x)P(x)
13+2(1)2−5(1)+k=01^3 + 2(1)^2 - 5(1) + k = 0
Evaluate the polynomial expression at x=1x = 1.
3
Simplify and solve for kk
k=2k = 2
Combine constants 1+2−5=−21 + 2 - 5 = -2 and solve −2+k=0-2 + k = 0.

Key Concept

Factor Theorem
Question 14Question →

If (2x−3)(2x - 3) is a factor of the polynomial P(x)=2x3−5x2+ax+6P(x) = 2x^3 - 5x^2 + ax + 6, what is the value of the constant aa?

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Answer: -1

Answer

The value of the constant aa is −1-1.
According to the Factor Theorem, if (2x−3)(2x - 3) is a factor of P(x)P(x), then P(32)=0P\left(\frac{3}{2}\right) = 0. Substituting x=32x = \frac{3}{2} gives 274−454+32a+6=0\frac{27}{4} - \frac{45}{4} + \frac{3}{2}a + 6 = 0, which simplifies to 32+32a=0\frac{3}{2} + \frac{3}{2}a = 0, resulting in a=−1a = -1.

Step-by-Step Solution

1
Apply the Factor Theorem by setting the linear factor equal to zero.
2x−3=0  ⟹  x=322x - 3 = 0 \implies x = \frac{3}{2}
If (2x−3)(2x - 3) is a factor of P(x)P(x), then P(32)=0P\left(\frac{3}{2}\right) = 0.
2
Substitute x=32x = \frac{3}{2} into the polynomial expression P(x)P(x).
P(32)=2(32)3−5(32)2+a(32)+6=0P\left(\frac{3}{2}\right) = 2\left(\frac{3}{2}\right)^3 - 5\left(\frac{3}{2}\right)^2 + a\left(\frac{3}{2}\right) + 6 = 0
Setting the resulting expression equal to zero allows solving for aa.
3
Simplify the powers and numerical terms.
2(278)−5(94)+32a+6=0  ⟹  274−454+32a+6=02\left(\frac{27}{8}\right) - 5\left(\frac{9}{4}\right) + \frac{3}{2}a + 6 = 0 \implies \frac{27}{4} - \frac{45}{4} + \frac{3}{2}a + 6 = 0
Evaluate each fraction before combining terms.
4
Combine constant terms and solve for aa.
−184+6+32a=0  ⟹  −92+6+32a=0  ⟹  32+32a=0  ⟹  a=−1-\frac{18}{4} + 6 + \frac{3}{2}a = 0 \implies -\frac{9}{2} + 6 + \frac{3}{2}a = 0 \implies \frac{3}{2} + \frac{3}{2}a = 0 \implies a = -1
Isolating aa yields the correct constant value.

Key Concept

Factor Theorem for linear divisors of the form (ax−b)(ax - b)
Question 15Question →

What is the remainder when the polynomial P(x)=x3−4x2+5x−2P(x) = x^3 - 4x^2 + 5x - 2 is divided by (x+1)(x + 1)?

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Answer: −12-12

Answer

−12-12
According to the Remainder Theorem, dividing a polynomial P(x)P(x) by a linear divisor (x−a)(x - a) yields a remainder equal to P(a)P(a). Here, the divisor is (x+1)(x + 1), which corresponds to a=−1a = -1. Substituting x=−1x = -1 into P(x)=x3−4x2+5x−2P(x) = x^3 - 4x^2 + 5x - 2 gives (−1)3−4(−1)2+5(−1)−2=−1−4−5−2=−12(-1)^3 - 4(-1)^2 + 5(-1) - 2 = -1 - 4 - 5 - 2 = -12. Thus, the value −12-12 is correct.

Step-by-Step Solution

1
Apply the Remainder Theorem
To find the remainder when P(x)P(x) is divided by (x+1)(x + 1), evaluate P(−1)P(-1) by setting x+1=0  ⟹  x=−1x + 1 = 0 \implies x = -1.
By the Remainder Theorem, dividing P(x)P(x) by (x−a)(x - a) leaves a remainder equal to P(a)P(a).
2
Substitute x=−1x = -1 into the polynomial P(x)=x3−4x2+5x−2P(x) = x^3 - 4x^2 + 5x - 2
P(−1)=(−1)3−4(−1)2+5(−1)−2P(-1) = (-1)^3 - 4(-1)^2 + 5(-1) - 2
Replace each instance of xx with −1-1.
3
Simplify the powers and terms
P(−1)=−1−4(1)−5−2=−1−4−5−2=−12P(-1) = -1 - 4(1) - 5 - 2 = -1 - 4 - 5 - 2 = -12
Compute arithmetic operations following standard order of operations.

Key Concept

Remainder Theorem
Estimated Time:45s
Question 16Question →

The polynomial P(x)=2x3+ax2+bx+6P(x) = 2x^3 + ax^2 + bx + 6 leaves a remainder of 1212 when divided by (x−1)(x - 1) and has (x+3)(x + 3) as a factor. What is the value of a−ba - b?

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Answer: 66

Answer

The value of a−ba - b is 66.
Applying the Remainder Theorem with P(1)=12P(1) = 12 gives a+b=4a + b = 4. Applying the Factor Theorem with P(−3)=0P(-3) = 0 gives 3a−b=163a - b = 16. Solving these simultaneous linear equations gives a=5a = 5 and b=−1b = -1. Evaluating a−b=5−(−1)a - b = 5 - (-1) gives 66.

Step-by-Step Solution

1
Apply the Remainder Theorem for the divisor (x−1)(x - 1)
a+b=4a + b = 4
According to the Remainder Theorem, P(1)=12P(1) = 12. Substituting x=1x = 1 gives 2(1)3+a(1)2+b(1)+6=12  ⟹  a+b+8=12  ⟹  a+b=42(1)^3 + a(1)^2 + b(1) + 6 = 12 \implies a + b + 8 = 12 \implies a + b = 4.
2
Apply the Factor Theorem for the factor (x+3)(x + 3)
3a−b=163a - b = 16
According to the Factor Theorem, P(−3)=0P(-3) = 0. Substituting x=−3x = -3 gives 2(−3)3+a(−3)2+b(−3)+6=0  ⟹  −54+9a−3b+6=0  ⟹  9a−3b=48  ⟹  3a−b=162(-3)^3 + a(-3)^2 + b(-3) + 6 = 0 \implies -54 + 9a - 3b + 6 = 0 \implies 9a - 3b = 48 \implies 3a - b = 16.
3
Solve the system of linear equations for aa and bb
a=5a = 5 and b=−1b = -1
Adding (a+b=4)(a + b = 4) and (3a−b=16)(3a - b = 16) yields 4a=20  ⟹  a=54a = 20 \implies a = 5. Substituting a=5a = 5 into a+b=4a + b = 4 gives b=−1b = -1.
4
Calculate the target value a−ba - b
a−b=6a - b = 6
Subtracting bb from aa gives 5−(−1)=5+1=65 - (-1) = 5 + 1 = 6.

Key Concept

Polynomial Remainder and Factor Theorems
Question 17Question →

When the polynomial P(x)=3x4−2x3+ax2+bx−12P(x) = 3x^4 - 2x^3 + ax^2 + bx - 12 is divided by (x2−4)(x^2 - 4), the remainder is 5x−45x - 4. What is the value of a+ba + b?

Show answer & explanation

Answer: 3

Answer

The value of a+ba + b is 33.
By using the Remainder Theorem for the quadratic divisor (x2−4)=(x−2)(x+2)(x^2 - 4) = (x - 2)(x + 2), evaluating P(2)=6P(2) = 6 gives 2a+b=−72a + b = -7, and evaluating P(−2)=−14P(-2) = -14 gives 2a−b=−332a - b = -33. Solving these linear equations simultaneously yields a=−10a = -10 and b=13b = 13, which sums to a+b=3a + b = 3.

Step-by-Step Solution

1
Set up the polynomial division relation.
P(x)=(x−2)(x+2)Q(x)+(5x−4)P(x) = (x - 2)(x + 2)Q(x) + (5x - 4)
By the Remainder Theorem and Division Algorithm, dividing by (x2−4)(x^2 - 4) yields a remainder of R(x)=5x−4R(x) = 5x - 4.
2
Find the values of P(2)P(2) and P(−2)P(-2) from the remainder.
P(2)=6P(2) = 6 and P(−2)=−14P(-2) = -14
Substituting the roots of the divisor x=2x = 2 and x=−2x = -2 eliminates the quotient term (x2−4)Q(x)(x^2 - 4)Q(x).
3
Substitute x=2x = 2 into the polynomial P(x)P(x) and set equal to 66.
2a+b=−72a + b = -7
3(16)−2(8)+4a+2b−12=20+4a+2b=6  ⟹  4a+2b=−143(16) - 2(8) + 4a + 2b - 12 = 20 + 4a + 2b = 6 \implies 4a + 2b = -14.
4
Substitute x=−2x = -2 into the polynomial P(x)P(x) and set equal to −14-14.
2a−b=−332a - b = -33
3(16)−2(−8)+4a−2b−12=52+4a−2b=−14  ⟹  4a−2b=−663(16) - 2(-8) + 4a - 2b - 12 = 52 + 4a - 2b = -14 \implies 4a - 2b = -66.
5
Solve the system of equations for aa and bb.
a=−10a = -10, b=13b = 13, and a+b=3a + b = 3
Adding the two linear equations gives 4a=−40  ⟹  a=−104a = -40 \implies a = -10. Substituting a=−10a = -10 into 2a+b=−72a + b = -7 gives b=13b = 13.

Key Concept

Polynomial Remainder Theorem for Non-Linear Divisors
Estimated Time:2m 30s
Question 18Question →

When the polynomial P(x)=x3−3x2+kx+12P(x) = x^3 - 3x^2 + kx + 12 is divided by (x−2)(x - 2), the remainder is 66. What is the value of the constant kk?

Show answer & explanation

Answer: -1

Answer

The value of the constant kk is −1-1.
By the Remainder Theorem, the remainder when P(x)P(x) is divided by (x−2)(x - 2) is P(2)P(2). Substituting x=2x = 2 into P(x)=x3−3x2+kx+12P(x) = x^3 - 3x^2 + kx + 12 gives P(2)=8−12+2k+12=8+2kP(2) = 8 - 12 + 2k + 12 = 8 + 2k. Setting 8+2k=68 + 2k = 6 yields 2k=−22k = -2, so k=−1k = -1.

Step-by-Step Solution

1
Apply the Remainder Theorem
P(2)=6P(2) = 6
Dividing P(x)P(x) by (x−2)(x - 2) leaves a remainder equal to evaluating P(x)P(x) at x=2x = 2.
2
Substitute x=2x = 2 into P(x)P(x) and set equal to 66
(2)3−3(2)2+2k+12=6(2)^3 - 3(2)^2 + 2k + 12 = 6
Set the evaluated polynomial equal to the given remainder.
3
Simplify the arithmetic terms
8+2k=68 + 2k = 6
Calculate powers and products: 8−12+12=88 - 12 + 12 = 8.
4
Solve the linear equation for kk
k=−1k = -1
Subtract 8 from both sides to get 2k=−22k = -2, then divide by 2.

Key Concept

Polynomial Remainder Theorem
Question 19Question →

Given that (x−2)(x - 2) is a factor of the polynomial P(x)=x3+kx2−5x+6P(x) = x^3 + kx^2 - 5x + 6, find the remainder when P(x)P(x) is divided by (x+3)(x + 3).

Show answer & explanation

Answer: -15

Answer

The remainder when P(x)P(x) is divided by (x+3)(x + 3) is −15-15.
According to the Factor Theorem, since (x−2)(x - 2) is a factor of P(x)=x3+kx2−5x+6P(x) = x^3 + kx^2 - 5x + 6, setting x=2x = 2 yields P(2)=0P(2) = 0. This gives 8+4k−10+6=08 + 4k - 10 + 6 = 0, which simplifies to 4k+4=04k + 4 = 0, giving k=−1k = -1. The polynomial is therefore P(x)=x3−x2−5x+6P(x) = x^3 - x^2 - 5x + 6. By the Remainder Theorem, dividing P(x)P(x) by (x+3)(x + 3) produces a remainder of P(−3)P(-3). Evaluating P(−3)=(−3)3−(−3)2−5(−3)+6=−27−9+15+6=−15P(-3) = (-3)^3 - (-3)^2 - 5(-3) + 6 = -27 - 9 + 15 + 6 = -15.

Step-by-Step Solution

1
Apply the Factor Theorem to determine the unknown constant kk.
k=−1k = -1
If (x−2)(x - 2) is a factor of P(x)P(x), then P(2)=0P(2) = 0. Substituting x=2x = 2 gives 23+k(2)2−5(2)+6=0  ⟹  4k+4=0  ⟹  k=−12^3 + k(2)^2 - 5(2) + 6 = 0 \implies 4k + 4 = 0 \implies k = -1.
2
Substitute k=−1k = -1 into the original polynomial to get the full expression.
P(x)=x3−x2−5x+6P(x) = x^3 - x^2 - 5x + 6
Replacing kk with −1-1 defines P(x)P(x) completely.
3
Apply the Remainder Theorem to find the remainder when P(x)P(x) is divided by (x+3)(x + 3).
Remainder is −15-15
By the Remainder Theorem, dividing P(x)P(x) by (x+3)(x + 3) leaves a remainder equal to P(−3)P(-3). Calculating P(−3)=(−3)3−(−3)2−5(−3)+6=−27−9+15+6=−15P(-3) = (-3)^3 - (-3)^2 - 5(-3) + 6 = -27 - 9 + 15 + 6 = -15.

Key Concept

Factor and Remainder Theorems for Polynomials
Estimated Time:1m 30s
Question 20Question →

When the polynomial P(x)=x3+ax2+bx−6P(x) = x^3 + ax^2 + bx - 6 is divided by (x−2)(x - 2), the remainder is 00. When P(x)P(x) is divided by (x+1)(x + 1), the remainder is 1212. What is the value of a+ba + b?

Show answer & explanation

Answer: −7-7

Answer

The value of a+ba + b is −7-7.
According to the Remainder Theorem, dividing P(x)P(x) by (x−2)(x - 2) with a remainder of 00 means P(2)=0P(2) = 0, giving the equation 2a+b=−12a + b = -1. Dividing P(x)P(x) by (x+1)(x + 1) with a remainder of 1212 means P(−1)=12P(-1) = 12, giving a−b=19a - b = 19. Solving these two linear equations simultaneously yields a=6a = 6 and b=−13b = -13. Adding these values together gives a+b=−7a + b = -7.

Step-by-Step Solution

1
Apply the Factor/Remainder Theorem for divisor (x−2)(x - 2)
P(2)=23+a(2)2+b(2)−6=0  ⟹  4a+2b+2=0  ⟹  2a+b=−1P(2) = 2^3 + a(2)^2 + b(2) - 6 = 0 \implies 4a + 2b + 2 = 0 \implies 2a + b = -1
Since dividing P(x)P(x) by (x−2)(x - 2) leaves a remainder of 00, P(2)=0P(2) = 0.
2
Apply the Remainder Theorem for divisor (x+1)(x + 1)
P(−1)=(−1)3+a(−1)2+b(−1)−6=12  ⟹  a−b−7=12  ⟹  a−b=19P(-1) = (-1)^3 + a(-1)^2 + b(-1) - 6 = 12 \implies a - b - 7 = 12 \implies a - b = 19
Setting the linear divisor x+1=0x + 1 = 0 gives x=−1x = -1, so P(−1)=12P(-1) = 12.
3
Solve the system of simultaneous linear equations for aa and bb
Adding (2a+b=−1)(2a + b = -1) and (a−b=19)(a - b = 19) yields 3a=18  ⟹  a=63a = 18 \implies a = 6. Substituting a=6a = 6 into a−b=19a - b = 19 gives 6−b=19  ⟹  b=−136 - b = 19 \implies b = -13.
Eliminating bb allows finding the values of constants aa and bb.
4
Calculate a+ba + b
a+b=6+(−13)=−7a + b = 6 + (-13) = -7
Summing the determined constants aa and bb gives the target expression.

Key Concept

Remainder and Factor Theorems
Estimated Time:1m 30s
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