3 mistakes learners make with factorising harder quadratics, each one named and explained.
With a leading coefficient, the pair of numbers has to multiply to a times c.
The factorisation is correct in form but one bracket still has a factor that could come out.
The brackets multiply back correctly, so nothing about the answer looks wrong.
Question. 6x^2 - 7x - 20
A common answer. (2x + 4)(3x - 5)
The answer. (3x + 4)(2x - 5)
Why. 2x + 4 still has a 2 in common, so this is not fully factorised, and it does not expand back correctly.
Scan each bracket for a shared factor before writing the answer down.
x squared minus 5x plus 6 is factorised with 1 and 6, which multiply to 6 but add to 7.
The product is the condition that is easier to test, so the first pair that works for it gets taken.
Question. x^2 - 5x + 6
A common answer. (x - 1)(x - 6), because 1 x 6 = 6
The answer. (x - 2)(x - 3)
Why. The pair has to do both jobs: multiply to 6 and add to -5, and only -2 and -3 do.
Two conditions, one pair. List the factor pairs and check the sum of each.
x squared minus 5x plus 6 = 0 is solved as x = -2 or x = -3.
The brackets say minus 2 and minus 3, and copying them out is the obvious next move.
Question. (x - 2)(x - 3) = 0
A common answer. x = -2 or x = -3
The answer. x = 2 or x = 3
Why. Each bracket has to equal zero, and x - 2 = 0 happens when x is +2.
Set each bracket to zero and solve it. The sign always turns over.
WAJD spots these patterns in your child's answers and names the one behind their wrong answers, instead of just marking them wrong.