Factor Completely
a.
x^2-5x-3
b.
x^2+23x+24
c.
6x^4-54x^2
d.
20x^2+100x+40
In general there are a number of ways of factoring. The last-ditch, but always working, method is to set the quadratic equal to zero and use the quadratic equation. This gives two solutions, x = a and x = b. Then the quadratic factors as (x - a)(x - b). (I call this a "last-ditch" method because it is the least elegant of the factoring methods.)Originally Posted by pashah
a) $\displaystyle x^2-5x-3$ This one just isn't going to factor nicely, so I'm going to use the quadratic formula to do this. The solutions to the quadratic, when set equal to zero are $\displaystyle \frac{5 \pm \sqrt{37}}{2}$. Thus
$\displaystyle x^2 - 5x - 3 = \left ( x - \frac{5 + \sqrt{37}}{2} \right ) \left ( x - \frac{5 - \sqrt{37}}{2} \right )$
Usually when we factor we factor over rational numbers, so I'm guessing there may be a typo here?
b) $\displaystyle x^2+23x+24$ This also is ugly and can be done using the quadratic formula.
c) $\displaystyle 6x^4-54x^2$ The common factor in both terms is $\displaystyle 6x^2$, so
$\displaystyle 6x^4-54x^2 = (6x^2) (x^2 - 9) = 6x^2(x+3)(x-3)$
(Using the difference of two squares factorization.)
d) $\displaystyle 20x^2+100x+40$ First factor out the common factor of 20:
$\displaystyle 20x^2+100x+40 = (20) (x^2 + 5x + 4)$.
We may factor the remaining quadratic by the "guess method." Since the coefficient of the x^2 term in $\displaystyle x^2 + 5x + 4$ is 1 we know that it factors as $\displaystyle (x - a)(x - b) = x^2 - (a+b)x + ab$. All we need do is guess at a and b.
So we know that ab = 4 and a + b = -5. Thus a and b are some combination of $\displaystyle \pm 1, 2, 4$ from the first equation and we get a = -1, b = -4 by trial and error from the second equation. Thus:
$\displaystyle x^2 + 5x + 4 = (x - 1)(x - 4)$. So we finally have:
$\displaystyle 20x^2+100x+40 = 20(x - 1)(x - 4)$.
-Dan
A standard way to see if a quadratic has "nice" factors is called the "ac method." Consider, for example, the problem of factoring the quadratic $\displaystyle 3x^2 + 11x - 4$ . (Given some practice at factoring you might be able to guess how this factors.)
Start by multiplying the coefficient of x^2 and the constant term: 3*-4 = -12. Now list all the factors of -12:
1, -12
2, -6
3, -4
4, -3
6, -2
12, -1.
Now find which of these pairs adds up to the coefficient of the x term: 11.
1 - 12 = -11
2 - 6 = -4
3 - 4 = -1
4 - 3 = 1
6 - 2 = 4
12 - 1 = 11 <-- This is it!
Now we use this in the original expression:
$\displaystyle 3x^2 + 11x - 4 = 3x^2 + (12x - 1x) - 4$
and we proceed by factoring by grouping:
$\displaystyle 3x^2 + 11x - 4 = 3x^2 + 12x - 1x - 4$ $\displaystyle = (3x^2 + 12x) + (-x - 4)$
$\displaystyle = 3x(x + 4) + -1(x+4) = (3x -1)(x+4)$.
It's kinda long, but it always works if the quadratic factors over the rational numbers. If you can't get it to work, that means that you need to use the quadratic formula to find the factors.
As a second example, let's do problem a) this way:
$\displaystyle x^2 - 5x - 3$.
So we multiply the coefficient of x^2 and the constant: 1*-3 = -3 and list all the factors of -3:
1, -3
3, -1.
Which of these pairs adds up to the coefficient of the x term: -5?
1 - 3 = -2
3 - 1 = 2.
None of them add to -5. Thus we need to use the quadratic formula, as I did in the previous post.
-Dan