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I got this question in college and I am unsure how to approach it. Determine the variable a such that the following expression can be simplified:

$\frac{x^2-4x + a}{x-7}$

Usually with these problems, I am able to factor out something from the denominator, but that isn't possible here. Instead in the numerator, I tried just write $x^2+4x$ like $x(x+4)$, but I am unsure where to go from here. If $a$ was $4$, I could write $x^2+4x+a$ as $(x-2)^2$, but the problem is I am supposed to find out a. Does anyone have any ideas how I can proceed in finding a such that the expression may be simplified?

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    $\begingroup$ Do you know how to do polynomial division? $\endgroup$ – Jam Sep 9 '16 at 12:17
  • $\begingroup$ @Jam not really. But how can I divide something here? There is nothing common in the numerator and the denominator. $\endgroup$ – user5846939 Sep 9 '16 at 12:19
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You want to be able to factorise $x^2-4x+a$ into $(x+b)(x-7)$ . Which is to say $$x^2+\bbox[lightgreen]{(b-7)}~x\bbox[gold]{-7b} = x^2\bbox[lightgreen]{-4}~x+\bbox[gold,2pt]{a}$$

So...

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You want to see whether $x^2-4x+a$ is divisible by $x-7$, which only happens if and only if $f(x)=x^2-4x+a$ has $7$ as a root. Since $$ f(7)=49-28+a=a+21 $$ you have $f(7)=0$ if and only if…

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$$\frac{x^2-4x+a}{x-7}=\frac{(x-7)^2+10(x-7)+21+a}{x-7}$$

to simplify it , the $21+a$ should be zero ,so the value of $a$ is $$a=-21$$

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  • $\begingroup$ Wow, thanks a lot. I have no idea how you did that though. Do you mind explaining the thought process you used to solve it? $\endgroup$ – user5846939 Sep 9 '16 at 12:25
  • $\begingroup$ $x^2-4x+a=x^2-14x+14x+49-49-4x+a=(x-7)^2+10x-49+a=(x-7)^2+10(x-7)+70-49+a$ $\endgroup$ – E.H.E Sep 9 '16 at 12:33
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$$y=\frac{x^2-4x+a}{x-7}$$

$$x^2-4x+a=x(x-7)+3(x-7)+(a+21)\frac{x-7}{x-7}$$

$$y=x+3+\frac{a+21}{x-7}$$

I'm on my phone so I can't see mistakes in algebra easily so let me know if I've explained something badly

The trick is to multiply the denominator of the two polynomial's, $(x-7)$, by whatever it needs for you to get the first term of the numerator $x^2$. Clearly we need to up the power by one so we multiply by $x$ to get $x\cdot(x-7)$. Now if we expanded that, we'd have $x^2-7x$ but we know our coefficient of $x$ needs to be $-4$. So we add $3(x-7)$ and continue like this until we've done all the terms of the polynomial on top. Voila.

If you're still a bit confused, look up polynomial division, there's many better explanations.

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by using the long division enter image description here

the remainder should be zero.

hence $$a=-21$$

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  • $\begingroup$ Maybe a stupid question, but why "should" the remainder be 0? $\endgroup$ – user5846939 Sep 9 '16 at 13:12
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    $\begingroup$ if the remainder not zero, the result of division will be $(x-7)(x+3)+\frac{a+21}{x-7}$ $\endgroup$ – E.H.E Sep 9 '16 at 13:14

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