Algebra Know-It-ALL

370 Quadratic Equations with Real Roots

Solution

The process of putting a quadratic into binomial factor form is called, not surprisingly, factoring. Learning

how to factor a quadratic takes some practice. Some quadratics are easy to factor, others are difficult, and

some can’t be factored at all—at least, not into binomials with real-number constants and coefficients.

Remember the general binomial factor form:

(px+q)(rx+s)= 0

We must find the values p,q,r, and s. We’re told that they’re all integers. The coefficient of the x^2 term in

the original equation is equal to 1. That means pr= 1 in the binomial factor form. The last constant in the

original equation is −15. This tells us that qs=−15. Because pr= 1, we have two possibilities:

• p= 1 and r= 1


• p=−1 and r=− 1

Because qs=−15, we have eight possibilities:

• q= 15 and s=− 1


• q=−15 and s= 1


• q= 1 and s=− 15


• q=−1 and s= 15


• q= 3 and s=− 5


• q=−3 and s= 5


• q= 5 and s=− 3


• q=−5 and s= 3

We can trim this down to four possibilities by eliminating duplicate scenarios. That leaves us with these

choices:

• q= 15 and s=− 1


• q=−15 and s= 1


• q= 3 and s=− 5


• q=−3 and s= 5

Now let’s look at the coefficient of x in the original equation. It’s −2. This tells us that

ps+qr=− 2

If we “play around” with our choices for awhile, we’ll see that if we let p= 1 ,q= 3 ,r= 1 , and s=−5, we get

ps+qr= 1 × (−5)+ 3 × 1

=− 5 + 3

=− 2

That’s the right coefficient for x in the original! Let’s try those numbers in the binomial factors for the left

side of the equation. We get

(px+q)(rx+s)= (x+ 3)(x− 5)

=x^2 − 5 x+ 3 x− 15

=x^2 − 2 x− 15