Factoring Quadratic x²+bx+c with Positive b

When tasked with factoring a polynomial of the form $x^2 + bx + c$, you must first find two numbers that equal $c$ when multiplied together and equal $b$ when added together.

For instance, let's say that we want to factor the polynomial $x^2 + bx + c$, and that we've found two numbers $d$ and $e$ such that $d ⋅ e = c$ and $d + e = b$.

We then rewrite our polynomial as $x^2 + dx + ex + c$, use parentheses to group $x^2 + dx$ and $ex + c$, and factor out greatest common factors from both those pairs of terms.

This should give us an expression that consists of two terms, both of which are being multiplied by the same binomial. Once we have factored out that binomial, we should be left with an expression consisting of two binomials that are being multiplied together: this is the factored form of our polynomial.

Example 1:

$$x^2 + 3x + 2$$

Since $2 ⋅ 1 = 2$ and $2 + 1 = 3$, we can rewrite our polynomial as follows:

$$x^2 + 2x +1x + 2$$

Now, we factor out greatest common factors from our polynomial's first two terms and from our polynomial's last two terms:

$$x(x + 2) +1(x + 2)$$

Finally, we factor the binomial $x + 2$ out of our expression (note that we are only able to do this because our expression consists of two terms that are both being multiplied by the binomial $x + 2$):

$$(x + 1)(x + 2)$$

Example 2:

$$x^2 + 10x + 25$$ $$= x^2 + 5x + 5x + 25$$ $$= x(x + 5) +5(x + 5)$$ $$= (x + 5)(x + 5)$$ $$= (x + 5)^2$$

Test Yourself!

This is an example of a perfect square trinomial!

Remember, this pattern doesn't hold for all quadratics. For example, there's a different pattern with a negative $b$ value!

(Video) Factoring

by mathman1024

This video by mathman1024 shows how to factor some basic trinomials.

$\large{\text{Summary}}$

While the general form of a trinomial is $ax^2+bx+c$, the most basic form of this expression is the quadratic $x^2+bx+c$, which we get when $a=1$.

Factoring an expression of the form $x^2+bx+c$ is essentially like reverse FOILing a pair of binomials that were previously multiplied together. For example, when we FOIL the expression $(x+5)(x+3)$, we get:

$$(x+5)(x+3)$$ $$\begin{array}{crcl} =&x^2&+3x+5x+&15\\ &\text{F}&\text{O I}&\text{L}\\ =&x^2&+8x+&15 \end{array}$$

On the other hand, if we wanted to factor the expression $x^2+8x+15$, we would perform the reverse of the process that we just walked through:

$$\begin{array}{crcl} &x^2&+8x+&15\\ =&x^2&+3x+5x+&15\\ \end{array}$$ $$=(x+5)(x+3)$$

The first expression that the video above shows us how to factor is $x^2+8x+15$. As our instructor notes, our goal in factoring this trinomial is to rewrite it as the product of two binomials:

$$x^2+8x+15\\ =(\ \ \ \ \ \ \ )(\ \ \ \ \ \ \ )$$

We begin factoring our trinomial by acknowledging that the expression's first term, $x^2$, came about from multiplying the first terms of our trinomial's two binomial factors. This means that, in order to figure out what the first terms of our trinomial's factors are, we need to find two terms that multiply to $x^2$; in other words, we need to ask ourselves how we can break the term $x^2$ into two pieces.

In this case, since both of our trinomial's factors must contain at least one term that consists of at least one variable, there are only two terms we can break our trinomial's $x^2$ into: $x$ and $x$.

We then plug those terms into our two empty sets of parentheses, like so:

$$x^2+8x+15\\ =(x\ \ \ \ \ \ \ )(x\ \ \ \ \ \ \ )$$

Next, in order to figure out what the last terms of our trinomial's factors are, we first must look at our trinomial's last term and write down all the pairs of terms that multiply to it (note: from the FOILing example that we walked through above, we can see that our trinomial's last term came from multiplying a $5$ and a $3$, but for the sake of this example, let's pretend that we don't know this piece of information yet).

In other words, we need to list all the factors of $15$:

$$\begin{array}{cc} 1&15\\ \hline 3&5 \end{array}$$

Now we know that the last terms of our trinomial's factors are either $1$ and $15$ or $3$ and $5$. However, only one of these combinations will give us our trinomial's middle term ($8x$) when the expression's factors are multiplied together.

When factoring a quadratic, we want to make sure that the last terms of the quadratic's factors add up to the coefficient of its middle term.

Since the coefficient of our trinomial's middle term is $8$, and since $3+5=8$, the only pair of terms from our list above that satisfies the aforementioned condition is $3$ and $5$ (note that, if we again look back to our FOILing example above, we see that our trinomial's middle term came from adding the terms $3x$ and $5x$).

Finally, we finish factoring our trinomial by plugging the terms $3$ and $5$ into our two now-completed sets of parentheses:

$$x^2+8x+15\\ =(x+3)(x+5)$$

After stepping us through the process of factoring the trinomial $x^2+8x+15$, our instructor shows us how to factor another trinomial: $x^2+12x+32$. Try factoring this expression on your own. Then, look to the video above to check your work.

Visual Factoring Model

Take notice of what the $b>0$ and $c>0$ mean. It describes these numbers are positive (they are greater than $0$.)

Let's model some factoring. $x^2+6x+8= (x+2)(x+4)$

When we factor, we see that $(x+4)(x+2)=x^2+6x+8$. Notice, when we factor a trinomial with only positive values, we get factors with only positive values. Let's now look at another model of factoring the expression $2x^2+7x+3$.

When we factor, we see that $(x+3)(2x+1)=2x^2+7x+3$.