Question

for what value of a, x-5 devide the polynomial x³-3x²+ax-10 exactly​

Answer 1

answer is 2522725381538253815390009

Answer 2

✎Question :-

For what value of a, x-5 divides the polynomial $\sf x^3-3x^2+ax-10$ exactly.

✎Solution :-

We have to find the value of a where $\sf x^3-3x^2+ax-10$ gets divided by x-5 exactly.

$\implies \sf x-5=0$

$\implies \sf x=5$

Substituting x = 5 in $\sf x^3-3x^2+ax-10$

$\implies \sf (5)^3-3(5)^2+a(5)-10=0$

$\implies \sf 125-75+5a-10=0$

$\implies \sf 40+5a=0$

$\implies \sf 5a = -40$

$\implies \sf a = \dfrac {-40}{5}$

$\implies \sf a= \dfrac{\cancel{-40}}{\cancel5}$

$\implies \sf a = -8$

✎Verification :-

$\sf a=-8$

We have to substitute this value along with $\sf x=5$ in $\sf x^3-3x^2+ax-10$ to verify :-

$\implies \sf (5)^3-3(5)^2+(-8)(5)-10=0$

$\implies \sf 125-75-40-10=0$

$\implies \sf 50-50=0$

$\implies \sf 0=0$

Hence, verified !!

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➽Therefore, for $\sf a = -8, x-5$ divides the polynomial $\sf x^3-3x^2+ax-10$.