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How to factor quadratic $ax^2+bx+c$?

If $x^2 + 2x - 35 = 0$, then $x = $?

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marked as duplicate by Qiaochu Yuan Jun 7 '11 at 10:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Please don't shout (all caps). Concerning your question: Why do you want to know/what have you tried? – t.b. Feb 9 '11 at 10:14
Ways to solve (or approximately solve) a quadratic include graphing it, using the quadratic formula, using the intermediate value theorem, using Newton's method, factoring the quadratic, and many more. Which techniques are you most comfortable with? – Joshua Shane Liberman Feb 9 '11 at 15:31

Well you can do this: Observe that $7 \times 5 =35$ and the difference between $7$ and $5$ is $2$, so you can write your equation as $$x^{2}+7x - 5x -35=0$$ which then can be written as $(x+7)\cdot (x-5)$.

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This is a quadratic equation. Try completing the square.

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Since one side of the equation is already zero, you use that $$AB = 0 \Longrightarrow A = 0 \text{ or } B = 0.$$ This reduces your problem to factoring the left hand side of the equation.

In general, when factoring monic quadratic polynomials, i.e., expressions of the form $x^2 + bx + c$, one wants to find two numbers that add to $b$ and multiply to $c$. In your case the numbers that add to $2$ and multiply to $-35$ are the numbers $7$ and $-2$. So,

$$x^2 + 2x - 35 = (x+7)(x-5).$$

Now, $(x+7)(x-5) = 0 \Longrightarrow (x + 7) = 0$ or $(x - 5) = 0$. This gives $x = -7$ or $x = 5$.

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