How do you factor $16 g^{2} - 72 g + 81$ $16g^2-72g+81$ ?

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Answer 1 · 2018-04-13T15:50:22

${(4 g - 9)}^{2}$ $(4g-9)^2$

$16 g^{2} - 72 g + 81 = {(4 g)}^{2} - 2 \cdot 4 g \cdot 9 + 9^{2}$ $16g^2-72g+81=(4g)^2-2 *4g*9+9^2$
$qquad = (4g-9)^2$

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