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For what values of a and b are x+1 and x+2 factors of x3 – ax2 – bx –8? Write down the other factor.

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For a = 1 and b = 10 x+1 and x+2 factors of x³-ax²-bx-8 = 0

Other factor is (x-4)

Explanation:

We have

x³-ax²-bx-8 = 0

Its factors are x+1 and x+2

That is x = -1 and x = -2 are its roots

Substituting x = -1

(-1)³-a(-1)²-b(-1)-8 = 0

-1 - a + b - 8 = 0

b - a = 9 ---------------------eqn 1

Substituting x = -2

(-2)³-a(-2)²-b(-2)-8 = 0

-8 - 4a + 2b - 8 = 0

2b - 4a = 16 ---------------------eqn 2

eqn 1 x -2

-2b + 2a = -18 ---------------------eqn 3

eqn 2 + eqn 3

-2a = -2

a = 1

Substituting in eqn 1

b - 1 = 9

b = 10

For a = 1 and b = 10, x+1 and x+2 factors of x³-ax²-bx-8 = 0

The equation is x³-x²-10x-8 = 0

Dividing with x + 1 we will get

x³-x²-10x-8 = (x+1)(x²-2x-8)

Dividing (x²-2x-8) with x + 2 we will get

x²-2x-8 = (x+2)(x-4)

So we have

x³-x²-10x-8 = (x+1)(x+2)(x-4)

Other factor is (x-4)

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