Find the remainder of $(2x^3-7x^2-19x+8)/(x^2-4x+5)$ without using division

Question

I have the following problem:

Find the remainder of $f(x)=2x^3-7x^2-19x+8$ is divided by $x^2-4x-5.$

An classmate said to equate coefficients, but I do not know what they referred to.

I have no idea how to proceed with this problem. Whether it's a hint or a full solution, any help would be appreciated.

I guess there are several such problems on this site itself. Here is one: http://math.stackexchange.com/q/873755/321264. — StubbornAtom, Sep 11 '16 at 18:25

score 11 · Answer 1 · answered Sep 11 '16 at 18:27

11

We have $f(x)=q(x)g(x)+r(x)$ with $\deg r<\deg g$. Hence $\deg(r)\le 1$. Now the hint: $$f(5)=q(5)g(5)+r(5)\text{ and }f(-1)=q(-1)g(-1)+r(-1).$$

answered Sep 11 '16 at 18:27

Hagen von Eitzen

2

5 and -1 chosen because they are the roots of $x^2 -4x -5$, in case that wasn't obvious. – Jeremy West Sep 11 '16 at 19:02

score 7 · Accepted Answer · answered Sep 11 '16 at 18:19

7

Hint

Find $a,b,c,d$ s.t. $$2x^3-7x^2-19x+8=(ax+b)(x^2-4x-5)+cx+d.$$

You immediately remark that $a=2$. I let you continue.

answered Sep 11 '16 at 18:19

Surb

2 Answers2