Question

Show that (x - 2), (x + 3) and (x - 4) are factors of x^3-3x^2-10x+24.​

Answer 1

Answer:

Given, (x−2) , (x+3) and (x−4) are factors of polynomial x

3

−3x

2

−10x+24.

Then, f(x)=x

3

−3x

2

−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.