# Math Help - Factorise the Following

1. ## Factorise the Following

Please can someone help in factorising the following quadratic using simple factors....

20x^2+7x-6

JP

2. Factoring $20x^2+ 7x- 6$ means finding (ax+ b)(cx+ d) so that the product, $acx^2+ (ad+ bc)x+ bd$ is equal to that. That means ac= 20, bd= -6, and ad+ bc= 7. It would be impossible to find a, b, c, and d without restricting them to integers (and might be impossible even so).

We are looking for integers a and b so that ab= 20. Well 20= 1*20, 2*10, 4*5, (-1)*(-20), (-2)*(-10), or (-4)*(-5). We are also looking for integers c and so that bd= -6. -6= (-1)*6, (-2)*3, 1*(-6), or 2*(-3). How do we decide which? TRY them! Which of those give ad+ bc= 7?

3. You can use the quadratic formula to factor. For $ax^2+bx+c=0$, $x={ \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}}$.

Factoring $20x^{2}+7x-6=0$.

$x={ \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}}= { \dfrac{-(7) \pm \sqrt{(7)^{2}-4(20)(-6)}}{2(20)}} = {\dfrac{-7 \pm 23}{40}}$.

Thus $x = \dfrac{2}{5}$ or $x = -\dfrac{3}{4}$. Now take each of these solutions:

$x = \dfrac{2}{5}$, multiplying both sides by 5 then subtracting 2 from both sides yields:

(1) $5x-2 = 0$.

$x = -\dfrac{3}{4}$, multiplying both sides by 4 and adding 3 to both sides yields:

(2) $4x+3=0$.

Multiplying equations (1) and (2) together we get the factorized form of $20x^{2}+7x-6=0$:

$(5x-2)(4x+3)=0$.