Question

show that (x-2) and (x-4) are factors of x3-3x2-10x+24.
answer carefully

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Answer 1

Given, (x−2) , (x+3) and (x−4) are factors of polynomial x^3−3x−10x+24.

Then, f(x)=x 3 −3x2 −10x+24.

If (x−2) is a factor, then x−2=0⟹x=2.

Replace x by 2, we get,

f(2)=(2)^3 −3(2)^2 −10(2)+24

f(2)=8−12−20+24

f(2)=0.

The value of f(2) is zero.

Then (x−2) is the factor of the polynomial x^3

−3x 2−10x+24.

If (x+3) is factor, then x+3=0⟹x=−3.

Replace x by −3, we get,

f(−3)=(−3)

3 −3(−3) 2 −10(−3)+24

f(−3)=−27−27+30+24

f(−3)=0.

The value of f(−3) is zero.

Then (x+3) is the factor of the polynomial x^3

−3x 2 −10x+24.

If (x−4) is factor, then x−4=0⟹x=4.

Replace x by 4, we get,

f(4)=(4)

3 −3(4) 2 −10(4)+24

f(4)=64−48−40+24

f(4)=0.

The value of f(4) is zero.

Then (x+4) is the factor of the polynomial x^3 −3x 2 −10x+24.

Therefore, hence showed that (x−2) , (x+3) and (x−4) are factors of polynomial x

3 −3x 2 −10x+24.

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Answer 2

Answer:

explained down

Step-by-step explanation:

Let the polyomial p(x)=x³-3x²-10x+24

= x³+(-2x²-x²)+(2x-12x)+24

=(x³-2x²)+(-x²+2x)+(-12x+24)

=x²(x-2)-x(x-2)-12(x-2)

=(x-2)(x²-x-12)

=(x-2)(x²-4x+3x-12)

=(x-2)[x(x-4)+3(x-4)]

=(x-2)[(x-4)(x+3)]

=(x-2)(x-4)(x+3)

Therefore,

Factors of x³-3x²-10x+24

= (x-2)(x-4)(x+3)