# Chapter 5 - Polynomials and Factoring - 5.4 Factoring Perfect-Square Trinomials and Differences of Squares - 5.4 Exercise Set: 106

$t(0.9+t)(0.9-t)$

#### Work Step by Step

Factoring the $GCF= t$, the given expression, $0.81t-t^3 ,$ is equivalent to \begin{array}{l} t(0.81-t^2) .\end{array} Using $x^2-y^2=(x+y)(x-y)$ or the factoring of the difference of 2 squares, then the factored form of the expression, $t(0.81-t^2),$ is \begin{array}{l} t(0.9+t)(0.9-t) .\end{array}

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