Question

What are the restrictions for a? 2a^2+a-15/5a^2+16a+3

Answer 1

ANSWER

The restrictions are

$a\ne -3,a\ne -\frac{1}{5}$

EXPLANATION

We were given the rational function,

$\frac{2a^2+a-15}{5a^2+16a+3}$

The function is defined for all values of a, except

$5a^2+16a+3=0$

This has become a quadratic trinomial, so we need to split the middle term.

We do that by multiplying the coefficient of $x^2$ which is 5 by the constant term which is 3. This gives us 15.

The factors of 15 that adds up to 16 are 1 and 15.

We use these factors to split the middle term.

$5a^2+15a+a+3=0$

We now factor to get,

$5a(a+3)+1(a+3)=0$

We factor further to get,

$(a+3)(5a+1)=0$

This implies that,

$(a+3)=0,(5a+1)=0$

This gives

$a=-3,a=-\frac{1}{5}$

These are the restrictions.