Question

Can (x – 5) be the remainder on division of a polynomial p(x) by (x + 8)?

Answer 1

polynomial p(x) is defined as

⇒p(x)=g(x)q(x)+r(x)

where g(x)= divisor ; q(x)= quotient and r(x)= remainder

∴ p(x) can be found by multiplying g(x) with q(x) & adding r(x) to the product.

(i).g(x)=(x−2); q(x)=x2−x+1; r(x)=4

∴p(x)=(x−2)[x2−x+1]+4

              =x3−x2+x−2x2+2x−2+4

              =x3−3x2+3x+2

(ii).g(x)=(x+3); q(x)=2x2+x+5; r(x)=3x+1

∴p(x)=(x+3)[2x2+x+5]+(3x+1)

              =2x3+x2+5x

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Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

DIVISION ALGORITHM

Let f(x), g(x) be two polynomials where g(x) # 0

Then there exists two polynomials q(x), r(x) such that

$p(x) = g(x)q(x) + r(x)$

Where

$r(x) = 0 \ \: : or \: \: degree \: r(x) < degree \: g(x)$

QUESTION

Can (x – 5) be the remainder on division of a polynomial p(x) by (x + 8)

EVALUATION

Here

$g(x) = x + 8$

So

$degree \: g(x) = 1$

$r(x) = x - 5$

$degree \: r(x) = 1$

Since

$degree \: r(x) = degree \: g(x)$

RESULT

So according to Division algorithm

(x – 5) can not be the remainder on division of a polynomial p(x) by (x + 8)