How do you factor $49 x^{2} - 144$ $49x^2-144$ ?

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How do you factor $49 x^{2} - 144$ $49x^2-144$ ?

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Answer 1 · 2018-05-14T23:38:29

(7x+12)(7x-12)

Explanation:

This problem is known as a 'perfect square.' The square root of 49 is 7 and the square root of $x^{2}$ $x^2$ is $x$ $x$ . The next number, 144, is also a perfect square and the answer to that is 12.
Firstly, multiply the first numbers/variables: 7 $x \cdot$ $x*$ 7 $x$ $x$ =49 $x^{2}$ $x^2$
Secondly, multiply the outer numbers/vars.: 7 $x \cdot$ $x*$ -12=-84 $x$ $x$
Thirdly, multiply the inner numbers/vars.: 7 $x \cdot$ $x*$ 12=84 $x$ $x$
Lastly, multiply the last numbers/vars.:12 $\cdot$ $*$ 12=144
The two number, 84 $x$ $x$ and 84 $x$ $x$ , cancel themselves out and leaves one with 49 $x$ $x$ -144
One CAN'T use this perfect square method IF the equation has a plus instead of a minus.

Answer 2 · 2018-05-14T23:55:42

${(7 x)}^{2} - 12^{2} = (7 x + 12) (7 - 12)$ $(7x)^2-12^2=(7x+12)(7-12)$

Explanation:

Factor:

$49 x^{2} - 144$ $49x^2-144$

This is a difference of squares. Use the formula:

$(a^{2} - b^{2}) = (a + b) (a - b)$ $(a^2-b^2)=(a+b)(a-b)$ ,

where:

$a = 7 x$ $a=7x$ and $b = 12$ $b=12$

Plug in the known values for $a$ $a$ and $b$ $b$ .

${(7 x)}^{2} - 12^{2} = (7 x + 12) (7 - 12)$ $(7x)^2-12^2=(7x+12)(7-12)$

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How do you factor $49 x^{2} - 144$ $49x^2-144$ ?

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