# How would I solve $(x + 7)(x - 3) = (x + 7)(5 - x)$?

I need to find the two roots (x-intercepts) from the equation:

$$(x + 7)(x - 3) = (x + 7)(5 - x)$$

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By inspection 7 is an x intercept. Now divide by (x+7) on both sides (it wont be 0 at the other intercept). This leaves you with x-3 = 5-x or 2x = 8, that is x= 4. – yjj Dec 7 '10 at 22:40
@yjj: perhaps post that as an answer? – Cam Dec 7 '10 at 22:42
@yjj: That should read -7. – Joe Johnson 126 Dec 7 '10 at 22:49
@Joe Johnson: Indeed... – yjj Dec 7 '10 at 22:55
Please use more descriptive question titles. – Rahul Dec 7 '10 at 23:09

Generally speaking, in an equation like yours with polynomial-type expressions (even in factored form, as yours are) on both sides, you probably want to collect everything on one side of the equals sign so that you have some polynomial-type expression equal to zero. You might do this by multiplying out both sides, then collecting all the terms on one side; or you might rewrite your equation $A=B$ as $A-B=0$ and factor the $A-B$ part, since your $A$ and $B$ are already factored.

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HINT $\rm\ \ A\ B\ =\ A\ C\ \iff\ A\ (B-C)\ =\ 0\ \iff\ A=0\ \ or\ \ B-C = 0$

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Hint: If $x+7\neq 0$, your equation is equivalent to $x-3=5-x$, because you can divide both sides by a quantity different from zero. And what does happen to your equation when $x+7=0$?

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So we want to solve the following equation: $(x+7)(x-3)=(x+7)(5-x)$.

$\underline{First~Step:}~~$ Distribute out terms on both sides of equal sign as follows:

$$x^{2}+4x-21=-x^{2}-2x+35$$

$\underline{Second~Step:}~~$ Subtract $(-x^{2}-2x+35)$ from both sides to give:

$$2x^{2}+6x-56=0$$

$\underline{Third~Step:}~~$ Multiply or divide entire equation by $\dfrac{1}{2}$ or $2$ respectively to give us:

$$x^{2}+3x-28=0$$

$\underline{Fourth~Step:}~~$From here we can factor this polynomial as the following factors:

$$(x-4)(x+7)=0$$ Leading us to our solution below,

$$x=4,~~x=-7$$

I hope this helps out. Let me know if there is something you need to be clarified a bit further. Thanks.

Good~Luck.

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No need to distribute out. – lhf Apr 20 '11 at 11:59
@lhf: I was not trying to take shortcuts, but rather let the OP see how the process can be done in a step-by-step fashion. I know that there are obvious shortcuts, but I did not want to take that road as the question asked "please show a step-by-step explanation". – night owl Apr 21 '11 at 12:04