How do you factor completely 14a2−68a+48?

1 Answer
MathFact-orials.blogspot.com
Jan 28, 2017

14a2−68a+48=2(7a−6)(a−4)

Explanation:

We have:

14a2−68a+48

We can first factor out the Largest Common Factor of 14, 68, and 48:

  • 14=2×7
  • 68=2×2×17
  • 48=2×2×2×2×3

The Largest Common Factor is 2:

2(7a2−34a+24)

Now we look for factors in the form of (ax+b)(cx+d) where

  • ac=7
  • bd=24
  • ad+bc=−34

We know the factors for 7=7×1, so let's set a=7,c=1

The factors for 24=−1×−24,−2×−12,−3×−8,−4×−6 (because ad+bc is a negative number and we've set ac to be positive, we need these factors to be negative).

Let's try some factors and trial and error our way into this:

a00000b0000c00000d000000ad00000bc000ad+bc

7000−80001000−3000−21000−8000−29

7000−60001000−4000−28000−6000−34

This gives us:

2(7a−6)(a−4)

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