If $x^2+bx+b$ is a factor of $x^3+2x^2+2x+c$, $c \neq 0$, then $b-c=?$

Question

I am not getting where to start! Please help me out

Welcome to math SE. People here like to see the efforts you made trying to solve your question. Please edit your post to include what you have tried so far. — Alain Remillard, Feb 18 '20 at 17:08
Welcome to MSE. In order for MathJax commands to take effect, they must be surrounded by $ signs. For example, $x^2$ is typeset as $x^2$ — saulspatz, Feb 18 '20 at 17:08
When in doubt, try polynomial long division. Carry $b$ and $c$ as parameters and then find values that make the remainder zero -- meaning, here, it has to be $0x+0$. You will find two solutions for $b$ and each one then gives a value for $c$, but one combination has $c=0$ and is not allowed by the problem statement. The other is the one that hits. — Oscar Lanzi, Feb 18 '20 at 17:10

mjw · Answer 1 · 2020-02-18T17:20:56.357

1

$$(x^2+bx+b)(x+\alpha)=x^3 + 2x^2 + 2x +c$$ $$x^3 + (b+\alpha)x^2 +b(\alpha+1)x + b\alpha = x^3 + 2x^2 +2x+c$$

So we have:

$$b+\alpha=2$$

$$b(\alpha+1)=2$$

$$b\alpha =c$$

Solving for $\alpha$ we get $\alpha(1-\alpha)=0$.

We cannot have $\alpha=0$ because then $c$ would be zero, so $\alpha=1$, $b=1$, $c=1$.

$$b-c=0.$$

edited Feb 18 '20 at 17:20

answered Feb 18 '20 at 17:11

mjw

Oh! Got it , thank u so much.god bless you – samarth jain Feb 18 '20 at 17:16
How do you get the last line? It does not hold when $c=0$ (then the difference is $2$) and I struggle to see how $c\not=0$ is incorporated into your development. – Oscar Lanzi Feb 18 '20 at 17:20
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@OscarLanzi, yes, there was a correction. – mjw Feb 18 '20 at 17:22

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