Question

How do you factor \[10{{x}^{2}}-7x-12\]?

Answer 1

Hint: We have the given expression which we have to factor. We have the quadratic equation in the form \[a{{x}^{2}}+bx+c\]. So, we will first multiply ‘a’ and ‘c’, and so we have \[-12\times 10=-120\]. We have to find zeros of the given equation such that the product of the zeroes is \[120\] and the sum of the zeroes is \[7\]. And we get the zeroes as \[-8\] and \[15\], so accordingly we will compute the equation and factorize the given equation completely.

Complete step by step solution:
According to the given question, we have an equation which we have to factorize.
So, the expression we are given is,
\[10{{x}^{2}}-7x-12\]----(1)
As we can see that the given equation is a quadratic equation of the form \[a{{x}^{2}}+bx+c\].
Product of zeroes \[=ac=\]\[10\times (-12)=-120\]
Sum of zeroes \[=b=7\]
So we will begin by finding the possible zeros of the given expression whose product gives the value as \[120\] and whose sum gives the value as 7.
We will write down the multiples of \[120\], and find the zeros whose sum give the value 7,
We have,
\[120=2\times 60=4\times 30=8\times 15\]
We can see that the numbers \[8\] and \[15\] satisfy the conditions to be the zeroes of the given quadratic equation. Their product gives the value \[120\] and when you subtract \[8\] from \[15\].
We get,
\[10{{x}^{2}}-7x-12\]
\[\Rightarrow 10{{x}^{2}}-(15-8)x-12\]
Opening up the brackets, we get,
\[\Rightarrow 10{{x}^{2}}-15x+8x-12\]
Taking out the common terms, we have,
\[\Rightarrow 5x(2x-3)+4(2x-3)\]
We can see that \[(2x-3)\] is common and so we get,
\[\Rightarrow (2x-3)(5x+4)\]
Therefore, the factorization of \[10{{x}^{2}}-7x-12=(2x-3)(5x+4)\].

Note: While finding the product and sum of the zeroes using the given equation, make sure that the signs are properly taken care of, else the factorization will not be correct. Also, while taking the common terms out, write the left out terms correctly.