the remainder when x cube + 3 PX + Q is divided by x minus a whole square is
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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
If it helps then mrk me as BRAINLIEST
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