How do you find all zeros of the function $f(x)=8x^2+53x-21$ ?

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Answer 1 · 2016-05-30T15:57:18

$x=3/8$ and $x=-7$

We can use an AC method to factor the quadratic.

Find a pair of factors of $AC=8*21 = 168$ which differ by $53$ .

The pair $56, 3$ works in that $56*3 = 168$ and $56-3 = 53$ .

Use this pair to split the middle term and factor by grouping:

$8x^2+53x-21$

$=8x^2+56x-3x-21$

$=(8x^2+56x)-(3x+21)$

$=8x(x+7)-3(x+7)$

$=(8x-3)(x+7)$

Hence the zeros of $f(x)$ are $x=3/8$ and $x=-7$

How do you find all zeros of the function f(x)=8x^2+53x-21f(x)=8x^2+53x-21?