# Factoring Trinomials

symbolipoint
Homework Helper
Gold Member
Why is this topic so long?
I see the original equation to solve: in pure text form, x(7x-8)=15

Distributive property is obvious to use, first.
7x^2-8x=15

7x^2-8x-15=0

Now if this is easily factorable, then try to set up these and test:
• (7x 3)(x 5)
• (7x 5)(x 3)
• (7x 1)(x 15)
• (7x 15)(x 1)

If any look good, check how the signs should work.

Mark44
Mentor
Why is this topic so long?
I see the original equation to solve: in pure text form, x(7x-8)=15

Distributive property is obvious to use, first.
7x^2-8x=15

7x^2-8x-15=0
The OP is trying to use the technique for quadratic trinomials of the form ##ax^2 + bx + c = 0##, in which you look for factors of ac that add up to b.

In this case, looking for factors of -105 that add up to -8. The pairs are (1, -105), (3, -35), (5, -21), and (7, -15). There are four more pairs, but all of them result in a positive sum.
In this case, the factors are 7 and -15, so you rewrite the equation as ##7x^2 + 7x - 15x - 15 = 0##, and then group them in pairs as ##7x(x + 1) - 15(x + 1) = 0##, resulting in ##(7x - 15)(x + 1) = 0##.

symbolipoint
Homework Helper
Gold Member
Thanks Mark44. Some of the discussion slipped by me because the topic is a little too long for what I saw in the original question posting.

symbolipoint
Homework Helper
Gold Member
The OP is trying to use the technique for quadratic trinomials of the form ##ax^2 + bx + c = 0##, in which you look for factors of ac that add up to b.

In this case, looking for factors of -105 that add up to -8. The pairs are (1, -105), (3, -35), (5, -21), and (7, -15). There are four more pairs, but all of them result in a positive sum.
In this case, the factors are 7 and -15, so you rewrite the equation as ##7x^2 + 7x - 15x - 15 = 0##, and then group them in pairs as ##7x(x + 1) - 15(x + 1) = 0##, resulting in ##(7x - 15)(x + 1) = 0##.
The second example should seem clear now.
You would have began with 7x^2-8x-105=0;
and then want to factor the quadratic expression.

See some two-factor combinations for just the 105.
105=5*21=7*15 and there are others.
....
I started into but did not think all of this through, thoroughly yet. I'll work it a few minutes and see.
...
Now I checked. NO combination for just integers worked. I would check with general solution for quadratic equation.

For sure, not factorable by integers. Discriminant is 3004=4*751, and the 751 is a prime number.

Last edited:
Mark44
Mentor
The second example should seem clear now.
You would have began with 7x^2-8x-105=0;
No, that equation isn't part of the problem. It arose from confusion on the part of the OP.
The equation in this problem is ##7x^2 - 8x - 15 = 0##.

His misunderstanding is the part about multiplying 7 and -15 (to get -105), and thinking that this product needs to show up in the equation. Here I've quoted what he wrote quite a few posts back.

Oh, gotcha. Isn't that what I did in my original post? I thought you are suppose to keep the -105 in the equation like this. ##7x^2 +7x | - 15x - 105 = 0##​

symbolipoint said:
and then want to factor the quadratic expression.

See some two-factor combinations for just the 105.
105=5*21=7*15 and there are others.
....
I started into but did not think all of this through, thoroughly yet. I'll work it a few minutes and see.
...
Now I checked. NO combination for just integers worked. I would check with general solution for quadratic equation.

For sure, not factorable by integers. Discriminant is 3004=4*751, and the 751 is a prime number.

SammyS