How do you factor $16 n^{2} + 43 n - 15$ $16n^2+ 43n -15$ ?

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How do you factor $16 n^{2} + 43 n - 15$ $16n^2+ 43n -15$ ?

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Answer 1 · 2016-05-31T14:31:15

$(16 n - 5) (n + 3)$ $(16n-5)(n+3)$

Explanation:

Any quadratic that can be factored can be expressed in the general form $(a x + b) (c x + d) = a c x^{2} + (a d + b d) x + b d$ $(ax+b)(cx+d) = acx^2 +(ad+bd)x +bd$

We therefore need to find factors of $16$ $16$ and $- 15$ $-15$ that in the right combination will give $43$ $43$

$16 = 16 \cdot 1$ $16 =16*1$ and $15 = 5 \cdot 3$ $15 =5*3$

$3 \cdot 16 = 48$ $3*16 = 48$

$48 - 5 \cdot 1 = 43$ $48-5*1=43$

$:.16n^2+43n-15 = (16n-5)(n+3)$

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How do you factor $16 n^{2} + 43 n - 15$ $16n^2+ 43n -15$ ?

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How do you factor 16n^2+ 43n -1516n2+43n−1516n^2+ 43n -15?

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How do you factor $16 n^{2} + 43 n - 15$ $16n^2+ 43n -15$ ?