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

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

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Answer 1 · 2016-05-30T15:57:18

$x = \frac{3}{8}$ $x=3/8$ and $x = - 7$ $x=-7$

Explanation:

We can use an AC method to factor the quadratic.

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

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

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

$8 x^{2} + 53 x - 21$ $8x^2+53x-21$

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

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

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

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

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

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

1 Answer

Explanation:

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Math

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How do you find all zeros of the function f(x)=8x^2+53x-21f(x)=8x2+53x−21f(x)=8x^2+53x-21?

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Explanation:

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