What is the remainder to x^3+4x^2+x-6 divided by (x-1)

Question

asked Apr 9, 2020 82.2k views

1 Answer

Nuwan · Answer 1 · 2020-04-13T16:04:54+0000

$x^3=x^2\cdot x$ , and
x^2(x-1)=x^3-x^2 . This gives a remainder of

(x^3+4x^2+x-6)-(x^3-x^2)=5x^2+x-6

$5x^2=5x\cdot x$ , and
5x(x-1)=5x^2-5x . This gives a new remainder of

(5x^2+x-6)-(5x^2-5x)=6x-6

$6x=6\cdot x$ , and
6(x-1)=6x-6 . This gives a new remainder of

(6x-6)-(6x-6)=0

and so there is no remainder.

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Quicker method: Use the polynomial remainder theorem, which says the remainder upon dividing a polynomial
p(x) by
x-c is
p(c) . Here we have

p(x)=x^3+4x^2+x-6

$c=1\implies p(c)=1+4+1-6=0$