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Polynomials, Factor and Remainder Theorems

22 soru

Tüm türler22Çoktan seçmeli14Sayısal8
Soru 21Soru →

Given that (x−3)(x - 3) is a factor of the polynomial P(x)=x3+kx2+mx−6P(x) = x^3 + kx^2 + mx - 6 and that dividing P(x)P(x) by (x+1)(x + 1) leaves a remainder of −12-12, what is the value of k+mk + m?

Cevabı ve açıklamayı göster

Cevap: −1-1

Cevap

The value of k+mk + m is −1-1.
Applying the Factor Theorem P(3)=0P(3) = 0 gives 3k+m=−73k + m = -7, and applying the Remainder Theorem P(−1)=−12P(-1) = -12 gives k−m=−5k - m = -5. Solving these simultaneous linear equations gives k=−3k = -3 and m=2m = 2, which sums to k+m=−1k + m = -1.

Adım Adım Çözüm

1
Apply the Factor Theorem for the linear factor (x−3)(x - 3)
3k+m=−73k + m = -7
By the Factor Theorem, if (x−3)(x - 3) is a factor, then P(3)=0P(3) = 0. Substituting x=3x = 3 gives 33+k(3)2+m(3)−6=0  ⟹  27+9k+3m−6=0  ⟹  9k+3m=−21  ⟹  3k+m=−73^3 + k(3)^2 + m(3) - 6 = 0 \implies 27 + 9k + 3m - 6 = 0 \implies 9k + 3m = -21 \implies 3k + m = -7.
2
Apply the Remainder Theorem for the divisor (x+1)(x + 1)
k−m=−5k - m = -5
By the Remainder Theorem, dividing P(x)P(x) by (x+1)(x + 1) gives remainder P(−1)=−12P(-1) = -12. Substituting x=−1x = -1 gives (−1)3+k(−1)2+m(−1)−6=−12  ⟹  −1+k−m−6=−12  ⟹  k−m−7=−12  ⟹  k−m=−5(-1)^3 + k(-1)^2 + m(-1) - 6 = -12 \implies -1 + k - m - 6 = -12 \implies k - m - 7 = -12 \implies k - m = -5.
3
Solve the system of linear equations for kk and mm
k=−3k = -3 and m=2m = 2
Adding the two equations (3k+m)+(k−m)=−7+(−5)(3k + m) + (k - m) = -7 + (-5) gives 4k=−12  ⟹  k=−34k = -12 \implies k = -3. Substituting k=−3k = -3 into k−m=−5k - m = -5 gives −3−m=−5  ⟹  m=2-3 - m = -5 \implies m = 2.
4
Calculate the target value k+mk + m
k+m=−1k + m = -1
Summing k=−3k = -3 and m=2m = 2 yields k+m=−3+2=−1k + m = -3 + 2 = -1.

Anahtar Kavram

Factor and Remainder Theorems
Tahmini Süre:1m 30s
Soru 22Soru →

Given that (x+2)(x + 2) is a factor of the polynomial P(x)=2x3−x2+ax+bP(x) = 2x^3 - x^2 + ax + b, and that dividing P(x)P(x) by (2x−1)(2x - 1) leaves a remainder of 1515, what is the value of a+ba + b?

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Cevap: 1414

Cevap

The value of a+ba + b is 1414.
Applying the Factor Theorem with x=−2x = -2 gives the equation −2a+b=20-2a + b = 20. Applying the Remainder Theorem with x=12x = \frac{1}{2} gives the equation a+2b=30a + 2b = 30. Solving these simultaneous equations yields a=−2a = -2 and b=16b = 16. Adding aa and bb gives 1414.

Adım Adım Çözüm

1
Apply the Factor Theorem for (x+2)(x + 2)
−2a+b=20-2a + b = 20
Since (x+2)(x + 2) is a factor of P(x)P(x), P(−2)=0P(-2) = 0. Substituting x=−2x = -2 yields 2(−2)3−(−2)2+a(−2)+b=02(-2)^3 - (-2)^2 + a(-2) + b = 0, which simplifies to −16−4−2a+b=0-16 - 4 - 2a + b = 0 or −2a+b=20-2a + b = 20.
2
Apply the Remainder Theorem for (2x−1)(2x - 1)
a+2b=30a + 2b = 30
Dividing P(x)P(x) by (2x−1)(2x - 1) leaves a remainder of 1515, so P(12)=15P\left(\frac{1}{2}\right) = 15. Substituting x=12x = \frac{1}{2} yields 2(18)−14+a2+b=152\left(\frac{1}{8}\right) - \frac{1}{4} + \frac{a}{2} + b = 15, which simplifies to a2+b=15\frac{a}{2} + b = 15 or a+2b=30a + 2b = 30.
3
Solve the system of linear equations for aa and bb
a=−2a = -2 and b=16b = 16
From step 1, b=2a+20b = 2a + 20. Substituting this into step 2 gives a+2(2a+20)=30  ⟹  5a+40=30  ⟹  5a=−10  ⟹  a=−2a + 2(2a + 20) = 30 \implies 5a + 40 = 30 \implies 5a = -10 \implies a = -2. Substituting a=−2a = -2 into b=2a+20b = 2a + 20 gives b=16b = 16.
4
Calculate a+ba + b
1414
Summing the calculated values gives a+b=−2+16=14a + b = -2 + 16 = 14.

Anahtar Kavram

Factor Theorem and Remainder Theorem for Polynomials
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Polynomials, Factor and Remainder Theorems Alıştırma Soruları — JAMB UTME — Sayfa 2 | Examkin