How do you factor $16{{h}^{2}}+8hk-15{{k}^{2}}$ ?

Hint: The given equation is a second degree equation of 2 variables, if we take one variable as constant, we can get a quadratic equation. So let’s take k as the constant number.so it is a quadratic equation of h. We know that we have to split the middle term to factorize a quadratic equation. So here we can write 8hk as 20hk – 12hk for factorization.

Complete step by step solution:
The given equation is $16{{h}^{2}}+8hk-15{{k}^{2}}$
If we assume k is a constant number and h is variable it is a quadratic equation with variable h. if we compare it to $a{{x}^{2}}+bx+c$ we get a = 16, b = 8k and c = $-15{{k}^{2}}$
The value of ac is equal to $-240{{k}^{2}}$ , we have to split 8hk such that product of coefficients of p is equal to $-240{{k}^{2}}$
We can write 8hk as 20hk – 12hk to factorize the equation
$\Rightarrow 16{{h}^{2}}+8hk-15{{k}^{2}}=16{{h}^{2}}+20hk-12hk-15{{k}^{2}}$
Now we can take 4h common from first 2 terms and – 3k common from last 2 terms
$\Rightarrow 16{{h}^{2}}+8hk-15{{k}^{2}}=4h\left( 4h+5k \right)-3k\left( 4h+5k \right)$
Now we can take 4h + 5k common from the whole equation.
$\Rightarrow 16{{h}^{2}}+8hk-15{{k}^{2}}=\left( 4h-3k \right)\left( 4h+5k \right)$

So $\left( 4h-3k \right)\left( 4h+5k \right)$ is the factor form of the equation $16{{h}^{2}}+8hk-15{{k}^{2}}$.

Note: If we can factorize a second- degree equation of 2 variables, then the equation represents a pair of straight lines in the cartesian plane. For example, the equation given in the question $16{{h}^{2}}+8hk-15{{k}^{2}}$ = 0 , if we assume h as y coordinate and k as x coordinate, the equation represent straight line 4y – 3x = 0 and 4y + 5x = 0.