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# Factorize: $9{{x}^{2}}-3x-20$ Verified
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Hint: We are given a quadratic equation and we are asked to factorize the equation which is given in terms of ‘x’. The given equation is of the form, $a{{x}^{2}}+bx+c$. So, we will be splitting the mid-term in order to factorize the quadratic equation. The mid-term should be split up such that the product of the split terms should be equal to $ac$ and their sum should be equal to $b$. As per our equation, the product is equal to -180 and their sum should be equal to -3, so we get our terms as -15 and 12. We will split up the middle term and get the common terms out. Hence, we will have factorized the given quadratic equation.

According to the given question, we are given a quadratic equation and we are asked to factorize the equation.
The equation that we have is,
$9{{x}^{2}}-3x-20$
We can see that the given equation is of the form $a{{x}^{2}}+bx+c$.
We will use the splitting middle term technique to factorize the equation.
The middle term should be such that the product of the terms should be equal to $ac=-180$ and the sum should be equal to $b=-3$
So, we can find it as,
\begin{align} & 180\times 1 \\ & 90\times 2 \\ & 45\times 4 \\ & 15\times 12 \\ \end{align}
That is, we can have the split terms as -15 and 12 as it follows the given condition.
So, we have,
$\Rightarrow 9{{x}^{2}}+(-15x+12x)-20$
$\Rightarrow 9{{x}^{2}}-15x+12x-20$
Now, we will pair up the terms and take the common terms out and we get the new expression as,
$\Rightarrow 3x(3x-5)+4(3x-5)$
Now, taking the common bracket out, we get,
$\Rightarrow (3x+4)(3x-5)$
Therefore, the factorization of the given quadratic equation is $(3x+4)(3x-5)$.

Note: The factorization should be carried out step wise without missing any terms. Also, pay attention to the signs of each of the terms in the given equation, as it is of significance during the process of factorization. The terms to be used while splitting up the middle term should be carefully found and with appropriate signs.