When the polynomial P(x) = 2x^4 + ax^3 + bx^2 - 5x + 6 is divided by (x - 2)(x + 1), the r… — JAMB UTME

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

According to the Remainder Theorem, dividing

P(x)

(x - 2)(x + 1)

leaves a remainder

R(x) = 6x + 8

. Substituting

x = -1

gives

R(-1) = 2

and

P(-1) = 2 - a + b + 5 + 6 = 13 - a + b

. Equating

13 - a + b = 2

gives

-a + b = -11

, which directly leads to

a - b = 11

Apply the Remainder Theorem for linear factors of the divisor

(x - 2)(x + 1)

Since the divisor is

(x - 2)(x + 1)

, the roots of the divisor are

x = 2

and

x = -1

. The remainder function is

R(x) = 6x + 8

, so

P(2) = R(2)

and

P(-1) = R(-1)

By the Polynomial Division Algorithm,

P(x) = (x - 2)(x + 1)Q(x) + R(x)

Evaluate

R(x)

and

P(x)

x = -1

R(-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 + b

.
Setting

P(-1) = R(-1)

gives

13 - a + b = 2

Substituting

x = -1

eliminates the quotient term since

(-1 + 1) = 0

Rearrange the equation to solve for

a - b

13 - a + b = 2 \implies -a + b = -11 \implies a - b = 11

Multiplying both sides of

-a + b = -11

-1

gives

a - b = 11

Remainder Theorem for Composite Linear Divisors

Soru