Factor polynomial $$ t^{5}-6t^{3}+9t $$

$$ t^{5}-6t^{3}+9t = t( t^{2}-3 )^{ 2 } $$

Step 1 :

After factoring out $ t $ we have:

$$ t^{5}-6t^{3}+9t = t ( t^{4}-6t^{2}+9 ) $$

Step 2 :

Both the first and third terms are perfect squares.

$$ x^4 = \left( \color{blue}{ t^{2} } \right)^2 ~~ \text{and} ~~ 9 = \left( \color{red}{ 3 } \right)^2 $$

The middle term ( $ -6x^2 $ ) is two times the product of the terms that are squared.

$$ -6x^2 = - 2 \cdot \color{blue}{t^{2}} \cdot \color{red}{3} $$

We can conclude that the polynomial $ t^{4}-6t^{2}+9 $ is a perfect square trinomial, so we will use the formula below.

$$ A^2 - 2AB + B^2 = (A - B)^2 $$

In this example we have $ \color{blue}{ A = t^{2} } $ and $ \color{red}{ B = 3 } $ so,

$$ t^{4}-6t^{2}+9 = ( \color{blue}{ t^{2} } - \color{red}{ 3 } )^2 $$

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