Mathematics

Mathematics

Problem

Let g(x)=2x3+ax2+bx−8g(x)=2x^{3}+ax^{2}+bx-8. The polynomial leaves a remainder 1010 when divided by x−2x-2 and is exactly divisible by x+3x+3. (a) Obtain two linear equations in aa and bb using the given information. (b) Solve these simultaneous equations to find the values of aa and bb. (c) Using the values of aa and bb, factorise g(x)g(x) completely. Provide complete working for each part.
Intuitive explanation

# Intuitive Explanation

The Remainder Theorem is your key tool here: when you divide a polynomial by (x−c)(x-c), the remainder equals g(c)g(c). Since g(x)g(x) leaves remainder 10 when divided by x−2x-2, you have g(2)=10g(2)=10; since it's exactly divisible by x+3x+3, you have g(−3)=0g(-3)=0. These two conditions give you two equations to solve for the unknowns aa and bb.

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# Complete Solution

## Part (a): Obtain two linear equations in aa and bb

USING THE REMAINDER THEOREM

When g(x)g(x) is divided by (x−2)(x-2), the remainder is g(2)g(2).

g(2)=2(2)3+a(2)2+b(2)−8=10g(2) = 2(2)^3 + a(2)^2 + b(2) - 8 = 10 16+4a+2b−8=1016 + 4a + 2b - 8 = 10 4a+2b+8=104a + 2b + 8 = 10 $$4a + 2b =

Intuitive explanation

# Intuitive Explanation

The Remainder Theorem is your key tool here: when you divide a polynomial by (x−c)(x-c), the remainder equals g(c)g(c). Since g(x)g(x) leaves remainder 10 when divided by x−2x-2, you have g(2)=10g(2)=10; since it's exactly divisible by x+3x+3, you have g(−3)=0g(-3)=0. These two conditions give you two equations to solve for the unknowns aa and bb.

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# Complete Solution

## Part (a): Obtain two linear equations in aa and bb

USING THE REMAINDER THEOREM

When g(x)g(x) is divided by (x−2)(x-2), the remainder is g(2)g(2).

g(2)=2(2)3+a(2)2+b(2)−8=10g(2) = 2(2)^3 + a(2)^2 + b(2) - 8 = 10 16+4a+2b−8=1016 + 4a + 2b - 8 = 10 4a+2b+8=104a + 2b + 8 = 10 $$4a + 2b =

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