Question

show that (x-2),(x+3),(x-4) are the factors of xcube-3xsquare-10x+24

Answer 1

Answer:

Step-by-step explanation:

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.

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