Example 1
Completion requirements
| Example 1 |
Determine the zeros of the function by factoring.
-
\(y = -2x^2 - 2x + 40 \)
Step 1: Identify the greatest common factor of all three terms.
GCF \(= -2\)
\(\begin{align}
y &= -2x^2 - 2x + 40 \\
&= -2\left( {x^2 + x - 20} \right) \\
\end{align}\)
Often when \(a \) is negative, a factor of \(–1\) is used as part of the GCF so the GCF has a negative sign and the coefficient for \(x^2\) is positive. Factoring out a negative value simply makes the rest of the factoring process easier. This is the convention that is almost always used.
Step 2: Factor the trinomial within the brackets.
Determine two numbers that multiply to \(-20\) and add to \(1\). The values \(-4\) and \(5\) work.
\(\begin{array}{l}
y = -2\left( {x^2 + x - 20} \right) \\
y = -2\left( {x - 4} \right)\left( {x + 5} \right) \end{array}\)
Step 3: Determine the zeros of the function.
\(\begin{align}
x - 4 = 0 \\
x = 4 \\
\end{align}\)\(\begin{align}
x + 5 = 0 \\
x = -5 \\
\end{align}\)
The zeros of the function are \(4 \) and \(–5\).
-
\(f(x) = 6x^2 + x - 12\)
Step 1: Identify the greatest common factor of all three terms.
There are no common factors.
Step 2: Factor \(f(x) = 6x^2 + x - 12\).
Determine two numbers that multiply to \(–72\) and add to \(1\). The two numbers that work are \(–8\) and \(9\). Decompose the middle term using coefficients of \(–8\) and \(9\); group the terms, and then factor.
\(\begin{array}{l}
f\left( x \right) = 6x^2 + x - 12 \\
f\left( x \right) = 6x^2 - 8x + 9x - 12 \\
f\left( x \right) = 2x\left( {3x - 4} \right) + 3\left( {3x - 4} \right) \\
f\left( x \right) = \left( {3x - 4} \right)\left( {2x + 3} \right) \end{array}\)
Step 3: Determine the zeros of the function.
\(\begin{align}
3x - 4 &= 0 \\
3x &= 4 \\
x &= \frac{4}{3} \\
\end{align}\)\(\begin{align}
2x + 3 &= 0 \\
2x &= -3 \\
x &= -\frac{3}{2} \\
\end{align}\)
The zeros of the function are \(\frac{4}{3}\) and \(-\frac{3}{2}\).