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I was doing math mentally, and I came upon a quadratic.

I have no experience in factoring quadratics mentally, so I'm wondering if there is a simple trick to factor quadratics mentally.

Question: Does anyone know a simple way to factor a quadratic mentally. Such as $6x^2+7x-3$

Frank
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  • You may find helpful the AC method. – Bill Dubuque Nov 15 '16 at 15:07
  • Using the ac method, or also known as grouping method, it becomes a matter of exercise to do these ones in the head. When you practice lots of those, you can do them in the head unless the numbers get too big. In your example, you can "see" that $(-2)(9)=-18$ and $-2+9=7$ – imranfat Nov 15 '16 at 15:10
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    Factors must be of the form $ax+b$ where $a=1,2,3,6$ and $b=\pm 1,3$. That's not too many to try in your head. Or use the quadratic formula to find roots - check first that the discriminant is a perfect square: $49 + 72 = 121$ in this case. – Ethan Bolker Nov 15 '16 at 15:11
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    @EthanBolker : I nearly always use your second suggestion. It's a good test to check if it's factorable in the first place (with rational coeffs). In this case, the roots of $f(x)=0$ are $\frac13$ and $-\frac32$ from which you can immediately see candidate factors of $3x-1$ and $2x+3$. +1 for you. – MPW Nov 15 '16 at 15:15

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Sorry if I'm late, but usually if a quadratic has the form $ax^2+bx+c$ it can be factored as $(qx-d)(px-f)$ where q, d, f, and p are some numbers where $f*d=c, q*p=a$. In this case, $6x^2+7x-3$, -3 has factors -1 and 3, and 6 has factors 3 and 2. Now we must check to find b. If the factor is 2x-1, then the other factor would be 3x+3, multiplying this out would give a b-value of 3, not seven, so the quadratic becomes $(3x-1)(2x+3)$ as desired. In general, factoring mentally takes practice, and there is no sure way to accomplish it. The AC-method works to an extent, but often times some quadratic equations cannot be factored using rational numbers. Often times however, the easiest method is the quadratic formula, which I personally find myself using often.