How do you factor $16 x^{2} + 8 x - 3$ $16x^2+8x-3$ ?

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How do you factor $16 x^{2} + 8 x - 3$ $16x^2+8x-3$ ?

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Answer 1 · 2016-10-02T22:17:24

$(4 x - 1) (4 x + 3)$ $(4x-1)(4x+3)$

Explanation:

If you cannot do this by inspection.
Then use the formula.
The solution to $y = a x^{2} + b x + c$ $y=ax^2+bx+c$ is $y = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $y=(-b+-sqrt(b^2-4ac))/(2a)$

So $y = \frac{- 8 \pm \sqrt{64 - 4.16 . - 3}}{32}$ $y=(-8+-sqrt(64-4.16.-3))/32$

Or $y = \frac{- 8 + 16}{32}$ $y=(-8+16)/32$ or $\frac{- 8 - 16}{32}$ $(-8-16)/32$
Or $y = \frac{8}{32}$ $y=8/32$ or $- \frac{24}{32}$ $-24/32$

Or $y = \frac{1}{4} or - \frac{3}{4}$ $y=1/4 or -3/4$

This gives the factorisation.....

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How do you factor $16 x^{2} + 8 x - 3$ $16x^2+8x-3$ ?

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How do you factor $16 x^{2} + 8 x - 3$ $16x^2+8x-3$ ?