Question

when p(x)=x cube -ax square +x is divided by (x+a) the remai
nder is ?

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Answer 1

Answer:

$-2a^3-a$

Step-by-step explanation:

$p(x)=x^3-ax^2+x$

Explaining Remainder Theorem

Let's suppose we have already divided p(x)  by $(x+a)$.

$p(x)=(x+a)Q(x)+R(x)$

Q(x) : quotient

R(x) : remainder

It's identity, because it is division.

Put $x=-a$ in the equation.

$p(-a)\\=0Q(-a)+R(-a)\\=R(-a)$

∴$p(-a)$ is the remainder.

$p(-a)\\=(-a)^3-a(-a)^2+(-a)\\=-a^3-a^3-a\\=-2a^3-a$

∴$-2a^3-a$