How do you find the product of ${(9 x + 4)}^{2}$ $(9x + 4)^2$ ?

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How do you find the product of ${(9 x + 4)}^{2}$ $(9x + 4)^2$ ?

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Answer 1 · 2017-04-12T00:49:38

$81 x^{2} + 72 x + 16$ $81x^2 + 72x + 16$

Explanation:

To simplify this expression, or any expression, a good start would be writing it out completely and getting rid of the exponent. This will make it a lot easier to multiply.

${(9 x + 4)}^{2}$ $(9x + 4)^2$ will become $(9 x + 4) (9 x + 4)$ $(9x + 4)(9x + 4)$

Now we can begin to multiply by using the FOIL method.

Take the $first term in the first binomial$ $color(blue)"first term in the first binomial"$ and multiply it with $every term in the second binomial$ $color(green)"every term in the second binomial"$ .

Then take the $second term in the binomial$ $color(red)"second term in the binomial"$ and multiply it with $every term in the second binomial$ $color(green)"every term in the second binomial"$ .

$(9 x + 4) (9 x + 4)$ $(9x + 4)(9x + 4)$

$(9 x + 4) (9 x + 4)$ $(color(blue)(9x) + 4)(color(green)(9x) + 4)$ $\to 9 x \cdot 9 x \to 81 x^{2}$ $color(orange)(->) 9x * 9x color(orange)(->) color(red)(81x^2)$

$(9 x + 4) (9 x$ $(color(blue)(9x) + 4)(9x$ $+ 4)$ $color(green)( + 4))$ $\to 9 x \cdot 4 \to 36 x$ $color(orange)(->) 9x * 4 color(orange)(->) color(red)(36x)$

$(9 x$ $(9x$ $+ 4) (9 x + 4)$ $color(red)( + 4))(color(green)(9x) + 4)$ $\to 4 \cdot 9 x \to 36 x$ $color(orange)(->) 4 * 9x color(orange)(->) color(red)(36x)$

$(9 x$ $(9x$ $+ 4) (9 x$ $color(red)( + 4))(9x$ $+ 4)$ $color(green)( + 4))$ $\to 4 \cdot 4 \to 16$ $color(orange)(->) 4 * 4 color(orange)(->) color(red)(16)$

Now all we have to do is add the terms that we got and simplify.

$81 x^{2} + 36 x + 36 x + 16$ $81x^2 + 36x +36x + 16$
$81 x^{2} + 72 x + 16$ $81x^2 + 72x + 16$

As you can see, when we simplify our initial expression, we get our answer which is $81 x^{2} + 72 x + 16$ $81x^2 + 72x + 16$ .

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How do you find the product of ${(9 x + 4)}^{2}$ $(9x + 4)^2$ ?

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How do you find the product of ${(9 x + 4)}^{2}$ $(9x + 4)^2$ ?