Question

big ideas math algebra 1: bts 2015 > chapter 7: factoring (x^2 + bx + c) > section exercises 7.5 > 23 factor (-6t - 16 + t^2). the factored expression is \boxed{}

Explanation:

Step1: Rearrange the expression

First, we rearrange the given expression \(-6t - 16 + t^{2}\) into standard quadratic form \(t^{2}-6t - 16\). This is because the standard form of a quadratic is \(ax^{2}+bx + c\), which makes factoring easier. So we have \(t^{2}-6t - 16\).

Step2: Factor the quadratic

We need to find two numbers that multiply to \(ac=- 16\) (where \(a = 1\) and \(c=-16\)) and add up to \(b=-6\). Let's list the factor pairs of \(-16\):

• \(1\times(-16)=-16\) and \(1+(-16)=-15\)
• \(2\times(-8)=-16\) and \(2+(-8)=-6\)

So the two numbers are \(2\) and \(-8\). We can rewrite the middle term using these numbers:
\(t^{2}+2t-8t - 16\)

Now, we factor by grouping. Group the first two terms and the last two terms:
\((t^{2}+2t)+(-8t - 16)\)

Factor out the greatest common factor (GCF) from each group:
From the first group, the GCF of \(t^{2}\) and \(2t\) is \(t\), so we get \(t(t + 2)\).
From the second group, the GCF of \(-8t\) and \(-16\) is \(-8\), so we get \(-8(t + 2)\).

Now our expression becomes:
\(t(t + 2)-8(t + 2)\)

We can factor out the common binomial factor \((t + 2)\):
\((t + 2)(t - 8)\)

Answer:

\((t - 8)(t + 2)\)