Question

the remainder when x cube + 3 PX + Q is divided by x minus a whole square is​

Answer 1

Step-by-step explanation:

Given p(x)=x^3+px +q->(1)

g(x)=(x-a)^2=x^2–2ax+a^2->(2)

As per Division algorithm,

x^3/x^2=x

x(x^2–2ax+a^2)=x^3–2ax^2+a^2x=q(x)->(3)

Now subtract (1)-(2)

0–2ax^2+px-a^2x+q=-2ax^2+x(p-a^2)+q->(4)

-2ax^2/x^2=-2a

-2a(x^2–2ax+a^2)=-2ax^2+4a^2x-2a^3->(5)

Subtract (4)-(5)

(p-a^2–4a^2)x-q+2a^3

(p-5a^2)x-q+2a^3

So remainder r(x)=(p-5a^2)x -q+2a^3

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