How do you factor $12 p^{2} + 16 p + 5$ $12p^2+16p+5$ ?

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Answer 1 · 2015-06-02T14:46:51

$(6 p + 5) (2 p + 1)$ $(6p+5)(2p+1)$

Use the quadratic formula for solving $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

Taking $b^{2} - 4 a c$ $b^2-4ac$ this is $16^{2} - 4.12 .5 = 256 - 240 = 16$ $16^2-4.12.5 = 256-240=16$ , $\sqrt{16} = 4$ $sqrt(16)=4$

This gives $p = \frac{- 16 \pm 4}{2.12}$ $p=(-16+-4)/(2.12)$ thus $p = - \frac{12}{24} or - \frac{20}{24}$ $p =-12/24 or -20/24$

Simplifying gives $p = - \frac{1}{2} or p = - \frac{5}{6}$ $p=-1/2 or p=-5/6$

These can be written as $2 p + 1 = 0 or 6 p + 5 = 0$ $2p+1=0 or 6p+5=0$

Hence $12 p^{2} + 16 p + 5 = (6 p + 5) (2 p + 1)$ $12p^2+16p+5=(6p+5)(2p+1)$

How do you factor 12p2+16p+512p^2+16p+5?