How do you factor $-6{{x}^{2}}-23x-20$ ?

Hint: We recall polynomials and its factorization. We recall the factorisation of quadratic polynomial $a{{x}^{2}}+bx+c$using splitting the middle term method where we find two number $p,q$ such that $p+q=b$ and $pq=c\times a$. We take $-1$ common from all the terms and then use splitting the middle term method. We find prime factorization of $c\times a$ to get $p,q$ and then take common necessary terms to factorize.

Complete step by step solution:
We know that when we divide a divided polynomial $p\left( x \right)$ with degree $n$ by some divisor polynomial $d\left( x \right)$ with degree $m\le n$ then we get the quotient polynomial $q\left( x \right)$ of degree $n-m$ and the remainder polynomial as $r\left( x \right)$.We use Euclidean division formula and can write as
\[p\left( x \right)=d\left( x \right)q\left( x \right)+r\left( x \right)\]
We also know that if the remainder polynomial is zero then we call $d\left( x \right),q\left( x \right)$ factor polynomial of $p\left( x \right)$ or simply factors of $p\left( x \right)$. We know that the general form quadratic polynomial is given by $a{{x}^{2}}+bx+c$. We factorize $a{{x}^{2}}+bx+c$ by splitting the middle term $b=p+q$ where $p\times q=c\times a$. We find $p,q$ as the factors from the prime factorization of $c\times a$. We are asked to find the factors of
\[-6{{x}^{2}}-23x-20\]
We take $-1$ common from all the terms in the above polynomial to have;
\[-\left( 6{{x}^{2}}+23x+20 \right)\]
We compare the above quadratic polynomial in brackets with general quadratic polynomials in one variable $a{{x}^{2}}+bx+c$ to have $a=6,b=23,c=20$. So we find $p,q$ such that $p+q=23,pq=6\times 20=120$. We prime factorize 120 to have
\[120=2\times 2\times 2\times 3\times 5\]
We see that if we take $p=2\times 2\times 2=8,q=3\times 5=15$ we get $pq=120$ and also $p+q=23$. So we can write the given polynomial as
\[-\left( 6{{x}^{2}}+8x+15x+20 \right)\]
We take $2x$ common from first two terms and 5 common from last two terms in the bracket to have;
\[\Rightarrow -\left( 2x\left( 3x+4 \right)+3\left( 3x+4 \right) \right)\]
We take $\left( 3x+4 \right)$ common from both terms to have;
\[\Rightarrow -\left( 3x+4 \right)\left( 2x+3 \right)\]
The above form is required factored form where factors are $-1,\left( 3x+4 \right),\left( 2x+3 \right)$.\[\]

Note: We note that if ${{p}_{1}}\left( x \right),{{p}_{2}}\left( x \right),...,{{p}_{k}}\left( x \right)$ are $k$ factors of $p\left( x \right)$ then we say $p\left( x \right)={{p}_{1}}\left( x \right){{p}_{2}}\left( x \right)...{{p}_{k}}\left( x \right)$ is factored completely if none of the factors ${{p}_{1}}\left( x \right),{{p}_{2}}\left( x \right),...,{{p}_{k}}\left( x \right)$ can be factored further. Here $-6{{x}^{2}}-23x-20=-\left( 3x+4 \right)\left( 2x+3 \right)$ is a complete factorization where $-1$ is the constant factor and the other two are linear factors.