How do you factor $x^{2} + 18 x + 81$ $x^2+18x+81$ ?

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How do you factor $x^{2} + 18 x + 81$ $x^2+18x+81$ ?

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Answer 1 · 2016-06-01T12:16:17

$x^{2} + 18 x + 81 = {(x + 9)}^{2}$ $x^2+18x+81=(x+9)^2$

Explanation:

As $81$ $81$ is the square of $9$ $9$ and middle term $18$ $18$ is double of this number, (compare it with ${(x + a)}^{2} = x^{2} + 2 a x + a^{2}$ $(x+a)^2=x^2+2ax+a^2$ )

Hence $x^{2} + 18 x + 81$ $x^2+18x+81$ is a complete square.

$x^{2} + 18 x + 81$ $x^2+18x+81$

= $x^{2} + 2 \times 9 \times x + 9^{2}$ $x^2+2xx9xx x+9^2$

= ${(x + 9)}^{2}$ $(x+9)^2$

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How do you factor $x^{2} + 18 x + 81$ $x^2+18x+81$ ?

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