How do you factor $10 x^{2} + 16 x - 55 x - 88$ $10x^2 + 16x - 55x - 88$ ?

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How do you factor $10 x^{2} + 16 x - 55 x - 88$ $10x^2 + 16x - 55x - 88$ ?

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Answer 1 · 2016-06-17T20:32:50

$(5 x + 8) (2 x - 11)$ $(5x + 8)(2x -11)$

Explanation:

Group the four terms into twp groups.

$(10 x^{2} + 16 x) + (- 55 x - 88)$ $(10x^2 + 16x) + (-55x -88)$

$2 x (5 x + 8) - 11 (5 x + 8) take out common factors$ $2x(5x +8) -11(5x + 8) " take out common factors"$

$(5 x + 8) (2 x - 11) common bracket as a factor$ $(5x + 8)(2x -11) " common bracket as a factor"$

Answer 2 · 2016-06-17T20:35:19

$(5 x + 8) (2 x + 1)$ $(5x+8)(2x+1)$

Explanation:

In the given polynomial we recognize having $2 x$ $2x$ as common factor between $10 x^{2} and 16 x$ $10x^2 and 16x$
Hence, $10 x^{2} + 16 x = 2 x (5 x + 8)$ $10x^2+16x=2x(5x+8)$

$- 11$ $-11$ is the common factor of $- 55 x and - 88$ $-55x and -88$

$- 55 x - 88 = - 11 (5 x + 8)$ $-55x-88=-11(5x+8)$

$10 x^{2} + 16 x - 55 x - 88$ $10x^2+16x-55x-88$
$= 2 x (5 x + 8) - 11 (5 x + 8)$ $=2x(5x+8)-11(5x+8)$
Taking the common factor $5 x + 8$ $5x+8$

$= (5 x + 8) (2 x - 11)$ $=(5x+8)(2x-11)$

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How do you factor $10 x^{2} + 16 x - 55 x - 88$ $10x^2 + 16x - 55x - 88$ ?

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How do you factor 10x2+16x−55x−8810x^2 + 16x - 55x - 88?

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How do you factor $10 x^{2} + 16 x - 55 x - 88$ $10x^2 + 16x - 55x - 88$ ?