How to Factor Trinomials?

Question

Can someone give me a text-only response on how to factor a trinomial?

I do not have the slightest clue. Please give me the steps using the formula $ax^2+bx+c$.

Answer 1

First, not every polynomial factors over integers (or rational numbers, or even the real numbers).

And since you are asking this question, I am going to assume that you are not ready for the quadratic formula.

Lets assume the of $a$ is positive. If it isn't, you can multiply through by $-1$

$ax^2 + bx + c = (px + r)(qx + s)$

Observations.
$pq = a\\ rs = c\\ ps + qr = b$

We will insist that $p,q$ are both positive. However, $r,s$ can be either positive or negative.

If $c$ is positive, then the sign $r,s$ must be the same.

If $c$ is positive, and $b$ is positive then $r,s$ are both positive.

If $c$ is positive, and $b$ is negative then $r,s$ are both negative.

If $c$ is negative, then one of $r,s$ is positive and one is negative.

$a, c$ only have finitely many factors. And if that number is fairly small, it is not too much work to just try out different pairs of factors as $p,q,r,s$ above, until something fits.

If that is frustrating you,

You can multiply $ac$ and factor that product. Either add factors or subtract factors using the rules of signs above, until you have a pair that gives the proper result for $b.$

And then you may have to factor those results until you find the correct values of $p,q,r,s$

Example:

$8x^2 + 6x - 35$

The sign for $c$ is negative. and $35 = 7\cdot 5$

You start with $(?x + 5)(?x - 7)$ or is it $(?x - 5)(?x + 7)?$

You don't know until you have tried some other numbers. And you take some guesses.

$(8x + 5)(x - 7), (4x + 5)(2x - 7)$ etc. until you find one that fits.

$(4x + 5)(2x - 7) = 8x^2 - 6x -35$ is close, but you have the sign flipped on the $b$ term.

$(4x - 5)(2x + 7) = 8x^2 + 6x -35$

Or you say:

$8\cdot (-35) = -280$

Find factor pairs of $-280$

$\begin {array} {}\\ 5&-56\\ 7&-40\\ 10&-28\\ 14&-20\\ 20&-14\\ 28&-10\\ \end{array}$

Now, we need a pair that sums to $6.$ And $(20,-14)$ fits the bill.

$ps = 20\\ qr = -14$

And $pq = 8 = 2\cdot4$

Since $4$ does not go into $14$

$p = 4, q = 2, s = \frac {20}4 = 5, r = \frac {-14}{2} = -7\\ (4x - 7)(2x + 5) = 8x^2 + 6x - 35$