MFG-tool Factors and \(x\)-Intercepts

Section 6.1 Factors and \(x\)-Intercepts

In this lesson we review the skills we need to solve quadratic equations by factoring.

Subsection 1. Multiply binomials

Subsubsection Examples

Example 6.1.

Expand the product \(~(2x+3)(x-6)~\text{.}\)

Multiply each term of the first binomial be each term of the second binomial. This gives four multiplications, often denoted by "FOIL," which stands for First terms, Outside terms, Inside terms, and Last terms.

\begin{align*} (2x+3)(x-6) \amp = \underline{x \cdot x}~+~\underline{2x \cdot (-6)}~+~\underline{(-3) \cdot x}~+~\underline{(-3) \cdot (-6)}\\ \amp \quad \quad \alert{\text{F}}\quad \quad \quad \quad \alert{\text{O}} \quad \quad \quad \quad \quad \alert{\text{I}} \quad \quad \quad \quad \quad \alert{\text{L}}\\ \amp = 2x^2-12x-3x+18 ~~~~~~~~\quad\blert{\text{Combine like terms.}}\\ \amp = 2x^2-15x+18 \end{align*}

Example 6.2.

Expand the product \(~-2(3x-4)(3x-5)\text{.}\)

Solution

First, multiply the binomial factors together.

\begin{equation*} (3x-4)(3x-5) = 9x^2-27x+20 \end{equation*}

Then use the distributive law to multiply the result by the monomial factor, \(-2\text{.}\)

\begin{equation*} \alert{-2}(9x^2-27x+20) = -18x^2+54x-40 \end{equation*}

Subsubsection Exercises

Checkpoint 6.3.

Expand the product \(~(2x+1)(3x-2)\text{.}\)

Answer

\(6x^2-x-2\)

Checkpoint 6.4.

Expand the product \(~(2t+5)(2t+5)\text{.}\)

Answer

\(4t^2+20t+25\)

Checkpoint 6.5.

Expand the product \(~4(a-3)(3a-5)\text{.}\)

Answer

\(12a^2-56a+60\)

Checkpoint 6.6.

Expand the product \(~-3(2b-3)(5b+1)\text{.}\)

Answer

\(-30b^2+39b+9\)

Subsection 2. Factor quadratic trinomials

To factor the trinomial \(x^2+bx+c\text{,}\) we look for two numbers \(p\) and \(q\) whose product \(pq\) is the constant term and whose sum \(p+q\) is the coefficient of the middle term.

\begin{equation*} \begin{aligned}[t] (x+p)(x+q) \amp = x^2+qx+px+pq\\ \amp = x^2 +\blert{(p+q)}x +\alert{pq} = x^2+\blert{b}x+\alert{c} \end{aligned} \end{equation*}

Sign Patterns for Quadratic Trinomials.

Assume that \(b,~c,~p\) and \(q\) are positive integers. Then

\(x^2+bx+c = (x+p)(x+q)\)

If all the coefficients of the trinomial are positive, then both \(p\) and \(q\) are positive.
\(x^2-bx+c = (x-p)(x-q)\)

If the middle term of the trinomial is negative and the other two terms are positive, then \(p\) and \(q\) are both negative.
\(x^2 \pm bx-c = (x+p)(x-q)\)

If the constant term of the trinomial is negative, then \(p\) and \(q\) have opposite signs.

Subsubsection Examples

Example 6.7.

Factor \(~t^2+7t+12~\) as a product of two binomials,

\begin{equation*} t^2+7t+12 = (t+p)(t+q) \end{equation*}

Solution

The constant term is 12, so we look for two numbers \(p\) and \(q\) whose product is 12. There are three possibilities:

\begin{equation*} \text{ 1 and 12, 2 and 6, or 3 and 4} \end{equation*}

Because the middle term is \(7t\text{,}\) we must have \(p+q=7\text{.}\) We check each possibility and find that \(p=3\) and \(q=4\text{.}\) Thus,

\begin{equation*} t^2+7t+12 = (t+3)(t+4) \end{equation*}

Example 6.8.

Factor \(~x^2-12x+20~\text{.}\)

Solution

For this example we must find two numbers \(p\) and \(q\) for which \(pq=20\) and \(p+q=-12\text{.}\) These two conditions tell us that \(p\) and \(q\) must both be negative. We start by listing all the ways to factor 20 with negative factors:

\begin{equation*} -1~~\text{and}~~-20,~~~~-2~~\text{and}~~-10,~~~~-4~~\text{and}~~-5 \end{equation*}

We check \(p+q\) for each possibility to see which one gives the correct middle term. Because \(-2+(-10) = -12\text{,}\) the factorization is

\begin{equation*} x^2-12x+20 = (x-2)(x-10) \end{equation*}

Example 6.9.

Factor \(~x^2+2x-15~\text{.}\)

Solution

This time the product \(pq\) must be negative, so \(p\) and \(q\) must have opposite signs, one positive and one negative. There are only two ways to factor 15, either 1 times 15 or 3 times 5. We just "guess" that the second factor is negative, and check \(p+q\) for each possibility:

\begin{equation*} 1-15=-14~~~~~~\text{or}~~~~~~3-5=-2 \end{equation*}

The middle term we want is \(2x\text{,}\) not \(-2x\text{,}\) so we change the signs of \(p\) and \(q\text{:}\) we use \(-3\) and \(+5\text{.}\) The correct factorization is

\begin{equation*} x^2+2x-15 = (x-3)(x+5) \end{equation*}

Subsubsection Exercises

Checkpoint 6.10.

Factor \(~x^2+8x+15\)

Answer

\((x+3)(x+5)\)

Checkpoint 6.11.

Factor \(~y^2+14y+49\)

Answer

\((y+7)(y+7)\)

Checkpoint 6.12.

Factor \(~m^2-10m+24\)

Answer

\((m-4)(m-6)\)

Checkpoint 6.13.

Factor \(~m^2-11m+24\)

Answer

\((m-3)(m-8)\)

Checkpoint 6.14.

Factor \(~t^2+8t-48\)

Answer

\((t+12)(t-4)\)

Checkpoint 6.15.

Factor \(~t^2-8t-48\)

Answer

\((t-12)(t+4)\)

Subsection 3. Solve quadratic equations

Subsubsection Examples

Example 6.16.

Solve \(~3x^2=48\)

Solution

We use extraction of roots. We first divide by 3 to isolate the squared expression.

\begin{align*} x^2 \amp = 16 \amp \amp \blert{\text{Take square roots.}}\\ x \amp = \pm 4 \end{align*}

The solutions are \(x=4\) and \(x=-4\text{.}\)

Example 6.17.

Solve \(~3x^2=12x\)

Solution

We solve by factoring. First, we get zero on one side of the equation.

\begin{align*} 3x^2-12x \amp = 0 \amp \amp \blert{\text{Factor the left side.}}\\ 3x(x-4) \amp = 0 \amp \amp \blert{\text{Set each factor equal to zero.}}\\ 3x=0,~~~~x-4 \amp = 0 \end{align*}

The solutions are \(x=0\) and \(x=4\text{.}\)

Example 6.18.

Solve \(~3x^2-10x-8=0\)

Solution

We solve by factoring. We factor the left side.

\begin{align*} (x-4)(3x+2) \amp = 0 \amp \amp \blert{\text{Set each factor equal to zero.}}\\ x-4=0,~~~~3x+2 \amp = 0 \amp \amp \blert{\text{Solve each equation.}}\\ x=4,~~~~x\amp = \dfrac{-2}{3} \end{align*}

The solutions are \(x=0\) and \(x=\dfrac{-2}{3}\text{.}\)