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