Please can someone help in factorising the following quadratic using simple factors....
20x^2+7x-6
Thanks in advance.
JP
Factoring $\displaystyle 20x^2+ 7x- 6$ means finding (ax+ b)(cx+ d) so that the product, $\displaystyle acx^2+ (ad+ bc)x+ bd$ 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?
You can use the quadratic formula to factor. For $\displaystyle ax^2+bx+c=0$, $\displaystyle x={ \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}}$.
Factoring $\displaystyle 20x^{2}+7x-6=0$.
Using the quadratic formula:
$\displaystyle x={ \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}}= { \dfrac{-(7) \pm \sqrt{(7)^{2}-4(20)(-6)}}{2(20)}} = {\dfrac{-7 \pm 23}{40}}$.
Thus $\displaystyle x = \dfrac{2}{5}$ or $\displaystyle x = -\dfrac{3}{4}$. Now take each of these solutions:
$\displaystyle x = \dfrac{2}{5}$, multiplying both sides by 5 then subtracting 2 from both sides yields:
(1) $\displaystyle 5x-2 = 0$.
$\displaystyle x = -\dfrac{3}{4}$, multiplying both sides by 4 and adding 3 to both sides yields:
(2) $\displaystyle 4x+3=0$.
Multiplying equations (1) and (2) together we get the factorized form of $\displaystyle 20x^{2}+7x-6=0$:
$\displaystyle (5x-2)(4x+3)=0$.