Question

How do you factor $9{{x}^{2}}-6x-15$ ?

Answer 1

Hint: When we factorize a quadratic equation $a{{x}^{2}}+bx+c$ we write $bx$ as $mx+nx$ where $m+n=b$ and $mn=ac$, then we can easily factorize the equation. We can split -6x to 9x and -15x to solve this question.

Complete step by step solution:
The given equation is $9{{x}^{2}}-6x-15$ which is a quadratic equation. if we compare the equation to standard quadratic equation $a{{x}^{2}}+bx+c$ then a = 9, b = -6 and c = -15
To factor a quadratic equation, we can find two numbers m and n such that the sum of m and n is equal to b and the product of m and n is $ac$. Then we can split $bx$ to $mx+nx$ then we can factor the equation easily.
In our case ac = -135 and b = -6
So pair of 2 numbers whose product is -135 and sum -6 is ( 9 ,-15)
We can -6x split to 9x – 15x
So $9{{x}^{2}}-6x-15=9{{x}^{2}}+9x-15x-15$
Taking 9x common in the first half of the equation and taking -15x common in the second half of the equation.
$\Rightarrow 9{{x}^{2}}-6x-15=9x\left( x+1 \right)-15\left( x+1 \right)$
Taking x + 1 common
$\Rightarrow 9{{x}^{2}}-6x-15=\left( 9x-15 \right)\left( x+1 \right)$
We can take 3 common from 9x – 15
$\Rightarrow 9{{x}^{2}}-6x-15=3\left( 3x-5 \right)\left( x+1 \right)$

Note:
While factoring a quadratic equation we can’t always split $bx$ such that their product is equal to ac because sometimes the roots can be irrational numbers. So in that case we can try the complete square method to factorize the equation.