I need to understand how to get from:
42x^2 - 9x - 6
to
3(2x-1)(7x+2)
sorry if it's in the wrong spot!
the systematic way would be
$42x^2 - 9x-6 = 42\left (x^2 - \dfrac{3}{14} - \dfrac 1 7\right)$
Now use the quadratic equation to find the roots of the quadratic in parens
$x = \dfrac{\dfrac{3}{14} \pm \sqrt{\dfrac{9}{196}+\dfrac 4 7}}{2} = \dfrac{\dfrac{3}{14}\pm \dfrac{11}{14}}{2} = \dfrac 1 2,~-\dfrac 2 7$
So then we have our original polynomial is given by
$42\left(x-\dfrac 1 2\right)\left(x + \dfrac 2 7\right) = 3(2x-1)(7x+2)$
$\displaystyle 3(14x^2 - 3x - 2) $
a*c method
14*(-2) = -28
You want two integer divisors of -28 which when
multiplied together give -28 and which add to -3.
-7 and 4 work.
Use them as coefficients on the first degree x-terms,
rewrite the polynomial, and then factor by grouping
on either of the next two versions that follow:
$\displaystyle 3[14x^2 + 4x - 7x - 2] \ \ \ or $
$\displaystyle 3[14x^2 - 7x + 4x - 2] \ =$
$\displaystyle 3[7x(2x - 1) \ \ \ ?(2x - 1)] \ =$
$\displaystyle 3[7x(2x - 1) \ + \ 2(2x - 1)] \ = $
$\displaystyle 3(2x - 1)(7x + 2)$