Question

factoring quadratics
factor each completely. find gcf first if needed.

1. $x^2 + x - 6$
2. $r^2 - 17r + 72$
3. $12n^2 - 76n + 120$
4. $7n^2 + 30n - 25$
5. $2r^2 - 18r$
6. $x^2 - 64$

Explanation:

Problem 1: \( x^2 + x - 6 \)

Step 1: Find two numbers

Find two numbers that multiply to \(-6\) (the constant term) and add up to \(1\) (the coefficient of \(x\)). The numbers are \(3\) and \(-2\) since \(3\times(-2)=-6\) and \(3 + (-2)=1\).

Step 2: Factor the quadratic

Rewrite the middle term using these two numbers: \(x^2 + 3x - 2x - 6\). Then factor by grouping: \(x(x + 3)-2(x + 3)=(x + 3)(x - 2)\).

Step 1: Find two numbers

Find two numbers that multiply to \(72\) (the constant term) and add up to \(-17\) (the coefficient of \(r\)). The numbers are \(-8\) and \(-9\) since \((-8)\times(-9)=72\) and \(-8 + (-9)=-17\).

Step 2: Factor the quadratic

Rewrite the middle term using these two numbers: \(r^2 - 8r - 9r + 72\). Then factor by grouping: \(r(r - 8)-9(r - 8)=(r - 8)(r - 9)\).

Step 1: Find GCF

First, find the greatest common factor (GCF) of \(12\), \(-76\), and \(120\). The GCF is \(4\). Factor out \(4\): \(4(3n^2 - 19n + 30)\).

Step 2: Factor the quadratic inside

Find two numbers that multiply to \(3\times30 = 90\) and add up to \(-19\). The numbers are \(-10\) and \(-9\). Rewrite the middle term: \(3n^2 - 10n - 9n + 30\). Factor by grouping: \(n(3n - 10)-3(3n - 10)=(3n - 10)(n - 3)\). Then multiply by the GCF: \(4(3n - 10)(n - 3)\).

Answer:

\((x + 3)(x - 2)\)

Problem 2: \( r^2 - 17r + 72 \)