How do you factor $40 x^{2} + 26 x - 6$ $40x^2 + 26x - 6$ ?

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How do you factor $40 x^{2} + 26 x - 6$ $40x^2 + 26x - 6$ ?

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Answer 1 · 2016-03-08T07:13:41

Factors are $40 (x - \frac{26 + 2 \sqrt{409}}{80}) (x - \frac{26 - 2 \sqrt{409}}{80})$ $40(x-(26+2sqrt409)/80)(x-(26-2sqrt409)/80)$

Explanation:

A simple way to factorize $a x^{2} + b x + c$ $ax^2+bx+c$ , one needs to split middle term $b x$ $bx$ in two components, so that their product is $a c$ $ac$ . However, this is not possible in $40 x^{2} + 26 x - 6$ $40x^2+26x−6$ , as product $a c$ $ac$ is $- 240$ $-240$ , $26 x$ $26x$ cannot be split into such set of integers.

Hence one ought to use quadratic formula. For this, let us find the roots of equation $40 x^{2} + 26 x - 6 = 0$ $40x^2+26x−6=0$ using $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b+-sqrt(b^2-4ac))/(2a)$

These are $\frac{- 26 \pm \sqrt{26^{2} - 4 \times 40 \times (- 6)}}{2 \times 40}$ $(-26+-sqrt(26^2-4xx40xx(-6)))/(2xx40)$ or

$\frac{- 26 \pm \sqrt{676 + 960}}{80}$ $(-26+-sqrt(676+960))/80$ or $\frac{- 26 \pm \sqrt{1636}}{80}$ $(-26+-sqrt(1636))/80$ or

$\frac{- 26 \pm 2 \sqrt{409}}{80}$ $(-26+-2sqrt409)/80$

Hence factors are $40 (x - \frac{26 + 2 \sqrt{409}}{80}) (x - \frac{26 - 2 \sqrt{409}}{80})$ $40(x-(26+2sqrt409)/80)(x-(26-2sqrt409)/80)$

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How do you factor $40 x^{2} + 26 x - 6$ $40x^2 + 26x - 6$ ?

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