When the polynomial P(x) = 3x^4 - 2x^3 + ax^2 + bx - 12 is divided by (x^2 - 4), the remai… — JAMB UTME

When the polynomial $P(x) = 3x^4 - 2x^3 + ax^2 + bx - 12$ is divided by $(x^2 - 4)$ , the remainder is $5x - 4$ . What is the value of $a + b$ ?

Answer: 3

The value of

a + b

3

By using the Remainder Theorem for the quadratic divisor

(x^2 - 4) = (x - 2)(x + 2)

, evaluating

P(2) = 6

gives

2a + b = -7

, and evaluating

P(-2) = -14

gives

2a - b = -33

. Solving these linear equations simultaneously yields

a = -10

and

b = 13

, which sums to

a + b = 3

Set up the polynomial division relation.

P(x) = (x - 2)(x + 2)Q(x) + (5x - 4)

By the Remainder Theorem and Division Algorithm, dividing by

(x^2 - 4)

yields a remainder of

R(x) = 5x - 4

Find the values of

P(2)

and

P(-2)

from the remainder.

P(2) = 6

and

P(-2) = -14

Substituting the roots of the divisor

x = 2

and

x = -2

eliminates the quotient term

(x^2 - 4)Q(x)

Substitute

x = 2

into the polynomial

P(x)

and set equal to

6

2a + b = -7

3(16) - 2(8) + 4a + 2b - 12 = 20 + 4a + 2b = 6 \implies 4a + 2b = -14

Substitute

x = -2

into the polynomial

P(x)

and set equal to

-14

2a - b = -33

3(16) - 2(-8) + 4a - 2b - 12 = 52 + 4a - 2b = -14 \implies 4a - 2b = -66

Solve the system of equations for

a

and

b

a = -10

b = 13

, and

a + b = 3

Adding the two linear equations gives

4a = -40 \implies a = -10

. Substituting

a = -10

into

2a + b = -7

gives

b = 13

Polynomial Remainder Theorem for Non-Linear Divisors

Estimated Time:2m 30s

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