Question

Find the zeros of the polynomial $x^2+4\sqrt{3}x-15$ by factorization method.

Answer 1

Hint: For solving the questions involving quadratic equations, we need to be aware about various terms related to the same. In this question, we are asked to find the zeros by factorization method. For this, we will try to find two numbers such that the product of those leads to the constant term in the expression. And the sum of both the numbers should be equal to the coefficient of $x$. Using such a technique would convert the equation in such a way that we would be able to take common from each term and we would be able to factorize the equation.

Complete step-by-step solution:
We have:
$x^2+4\sqrt{3}x-15$
We need to find zeros of this. The zeros of any quadratic equation means the numbers which when plugged in place of $x$ give 0 as the output. We need to factorize this equation now, so we find two numbers such that their product is the constant term i.e. -15 and the sum if the coefficient of $x$ i.e. $4\sqrt{3}$. In short we need to find $a$ and $b$ such that:
$a+b=-15$
$a\times b=4\sqrt{3}$
Now see that:
$5\sqrt{3}\times \sqrt{3}=5\times 3=15$
Also,
$5\sqrt{3}- \sqrt{3}=\left(5-1\right)\sqrt{3}=4\sqrt{3}$
So, we take $a=5\sqrt3$ and $b=\sqrt3$
And we dissolve the coefficient of $x$ as following:
$x^2+5\sqrt{3}x-\sqrt{3}x-15$
Now, we take the terms common to get:
$x^2+4\sqrt{3}x-15$
$=\left(x+5\sqrt{3}\right)\left(x-\sqrt{3}\right)$
Hence, we have found the factors of the quadratic equation which are $-5\sqrt{3}$ and $\sqrt{3}$.

Note: While finding numbers for factorization, you need to be sure that you don’t make any calculation mistakes because if you are not able to find the appropriate numbers then it might lead to an invalid result because then you won’t be able to factorize the quadratic equation further.