6.5.3 Factoring Trinomials Using Trial and Error

Factoring Trinomials Using Trial and Error

Our next step is to factor trinomials whose leading coefficient is not $1$, trinomials of the form $a{x}^2+bx+c$. Remember to always check for a GCF first! Sometimes, after you factor the GCF, the leading coefficient of the trinomial becomes $1$ and you can factor it by the methods we’ve used so far. Let’s do an example to see how this works.

Example 1

Factor completely: $4x^3+16x^2-20x$.

Step 1 - The trinomial is already written in descending order of degrees.

Step 2 - Is there a greatest common factor?

$4x^3+16x^2-20x$

Yes, $GCF=4x$. Factor it.

$4x(x^2+4x-5)$

Is it a binomial, trinomial, or more than three terms?

$4x(x^2+4x-5)$

It is now a trinomial with a leading coefficient of $1$. So “undo FOIL.”

$4x(x)(x)$

Step 3 - Use a table like the one shown to find two numbers that multiply to $-5$ and add to $4$.

Step 4 - Use −1 and 5 as coefficients of the last terms.

$4x(x-1)(x+5)$

Step 5 - Check.

$\begin{array}{c}4x\left(x-1\right)\left(x+5\right)\\ 4x(x^2+5x-x-5)\\ 4x(x^2+4x-5)\\ 4x^3+16x^2-20x\✓\end{array}$

What happens when the leading coefficient is not $1$ and there is no GCF? There are several methods that can be used to factor these trinomials. First we will use the trial and error method.

Example 2

Factor the trinomial $3x^2+5x+2$.

The trinomial is already written in descending order of degrees. There is no GCF to factor.

From our earlier work, we expect this will factor into two binomials.

$\begin{array}{c}\hfill 3x^2+5x+2\hfill \\ \hfill ()()\hfill \end{array}$

We know the first terms of the binomial factors will multiply to give us $3x{a}^2$. The only factors of $3x^2$ are $1x$, $3x$. We can place them in the binomials.

Check: Does  $1x·3x=3x^2$?

We know the last terms of the binomials will multiply to $2$. Since this trinomial has all positive terms, we only need to consider positive factors. The only factors of $2$ are $1$, $2$. But we now have two cases to consider because it will make a difference if we write $1$, $2$ or $2$, $1$.

Which factors are correct? To decide that, we multiply the inner and outer terms.

Since the middle term of the trinomial is $5x$, the factors in the first case will work. Let’s use FOIL to check.

$\begin{array}{c}\hfill (x+1)(3x+2)\hfill \\ \hfill 3x^2+2x+3x+2\hfill \\ \hfill 3x^2+5x+2✓\hfill \end{array}$

Our result of the factoring is:

$\begin{array}{c}\hfill 3x^2+5x+2\hfill \\ \hfill (x+1)(3x+2)\hfill \end{array}$

Remember, when the middle term is negative and the last term is positive, the signs in the binomials must both be negative.

When we factor an expression, we always look for a greatest common factor first. If the expression does not have a greatest common factor, there cannot be one in its factors either. This may help us eliminate some of the possible factor combinations.

Example 3

Factor completely using trial and error: $18x^2-37xy+15y^2$.

Step 1 - The trinomial is already in descending order.

$18x^2-37xy+15y^2$

Step 2 - Find the factors of the first term.

$$\begin{gathered} 18x^2-37xy+15y^2 \\ 1x·18x \\ 2x·9x \\ 3x·6x \end{gathered}$$

Step 3 - Find the factors of the last term. Consider the signs. Since 15 is positive and the coefficient of the middle term is negative, we use the negative factors.

Step 4 - Consider all the combinations of factors.

Step 5 - The correct factors are those whose product is the original trinomial.

$(2x-3y)(9x-5y)$

Step 6 - Check by multiplying.

$\begin{array}{c}\hfill \left(2x-3y)(9x-5y\right)\hfill \\ \hfill 18x^2-10xy-27xy+15y^2\hfill \\ \hfill 18x^2-37xy+15y^2\✓\hfill \end{array}$

In other trinomials, don’t forget to look for a GCF first, and remember if the leading coefficient is negative, so is the GCF.