Question

How do you factor $16{{p}^{2}}-4pq-30{{q}^{2}}$?

Answer 1

Hint: The given equation in the question is in 2 variables that are p and q. We can factorize the equation by assuming one variable is constant. Let's assume q is a constant number. Now the equation will be a quadratic equation where the variable is p. So we can write – 4pq as a sum of 2 terms where the product of coefficients is equal to $-480{{q}^{2}}$ .

Complete step by step solution:
The given equation is $16{{p}^{2}}-4pq-30{{q}^{2}}$
If we assume q is a constant number and p is variable it is quadratic equation with variable p. if we compare it to $a{{x}^{2}}+bx+c$ we get a = 16, b = - 4q and c = $-30{{q}^{2}}$
The value of ac is equal to $-480{{q}^{2}}$ , we have to split – 4pq such that product of coefficients of p is equal to $-480{{q}^{2}}$
We can write - 4pq as 20pq – 24 pq to factorize the equation
$\Rightarrow 16{{p}^{2}}-4pq-30{{q}^{2}}=16{{p}^{2}}+20pq-24pq-30{{q}^{2}}$
Now we can take 4p common from first 2 terms and - 6q common from last 2 terms
$\Rightarrow 16{{p}^{2}}-4pq-30{{q}^{2}}=4p\left( 4p+5q \right)-6q\left( 4p+5q \right)$
Now we can take 4p + 5q common from the whole equation.
$\Rightarrow 16{{p}^{2}}-4pq-30{{q}^{2}}=\left( 4p-6q \right)\left( 4p+5q \right)$

So $\left( 4p-6q \right)\left( 4p+5q \right)$ is the factor form of the equation $16{{p}^{2}}-4pq-30{{q}^{2}}$

Note: In the given question, there are 2 unknown variables and one equation. If we try to solve for the variable, we will only get one variable in terms of another variable so there will be infinitely many solutions for the equation. For factorization we can assume one variable as a constant term.