Expression Factoring?

Question

Can these even be factored or is there no factor?
1.x^2+3x-36
2.x^2-5x+82
3.6x^2-5x+876
4. 6a^2 +34a -122

Answer 1

To determine if a question factors, for a general quadratic expression ax^2 + bx + c, just test b^2 - 4ac. If you get a perfect square, then the expression is factorable.

1) x^2 + 3x - 36

b^2 - 4ac = 3^2 - 4(1)(-36) = 9 + 144 = 153.
153 is not a perfect square.
144 < 153 < 169, so
sqrt(144) < sqrt(153) < sqrt(169)
12 < sqrt(153) < 13

In theory these *can* be factored, but not in a simple way. The quadratic formula will give you two solutions; let's call them m and n. Once you know m and n are solutions, it follows that
(x - m) and (x - n) are factors.

x^2 + 3x - 36 = 0 implies
x = [-b +/- sqrt(b^2 - 4ac)] / (2a)
x = [-3 +/- sqrt(9 - 4(1)(-36)] / 2
x = [-3 +/- sqrt(9 + 144)] / 2
x = [-3 +/- sqrt(153)] / 2
x = [-3 +/- 3sqrt(17)] / 2
x = (-3/2) )(1 +/- sqrt(17)

Let m = (-3/2) (1 + sqrt(17)) and
n = (-3/2) (1 - sqrt(17))

Then x^2 + 3x - 36 factors into (x - m)(x - n)