Question

question 11
rewrite each of the following polynomial expressions in factored form. (make sure it is completely factored!)
$-5y^2 + 80 = square$
$5y^2 - 25y + 20 = square$
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Explanation:

Step1: Factor out -5 from the first expression

First, we factor out the greatest common factor (GCF) from \(-5y^{2}+80\). The GCF of \(-5y^{2}\) and \(80\) is \(-5\). So we have:
\(-5y^{2}+80=-5(y^{2} - 16)\)
Now, notice that \(y^{2}-16\) is a difference of squares, since \(y^{2}-16=y^{2}-4^{2}\). The formula for factoring a difference of squares is \(a^{2}-b^{2}=(a + b)(a - b)\). Here, \(a = y\) and \(b = 4\), so \(y^{2}-4^{2}=(y + 4)(y - 4)\). Therefore, \(-5y^{2}+80=-5(y + 4)(y - 4)\)

Step2: Factor the second expression

For the expression \(5y^{2}-25y + 20\), first factor out the GCF, which is \(5\):
\(5y^{2}-25y + 20=5(y^{2}-5y + 4)\)
Now, we need to factor the quadratic \(y^{2}-5y + 4\). We need two numbers that multiply to \(4\) and add up to \(-5\). The numbers are \(-1\) and \(-4\) because \((-1)\times(-4)=4\) and \(-1+(-4)=-5\). So we can factor \(y^{2}-5y + 4\) as \((y - 1)(y - 4)\). Therefore, \(5y^{2}-25y + 20=5(y - 1)(y - 4)\)

Answer:

For \(-5y^{2}+80\), the factored form is \(\boldsymbol{-5(y + 4)(y - 4)}\)

For \(5y^{2}-25y + 20\), the factored form is \(\boldsymbol{5(y - 1)(y - 4)}\)