What is the factorization of the polynomial below? -x^2-15x-56

Answer:
Option C. [tex](x+8)(x+7)[/tex]
Step-by-step explanation:
we have
[tex]-x^2-15x-56[/tex]
Find the roots of the quadratic equation
equate the equation to zero
[tex]-x^2-15x-56=0[/tex]
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]-x^2-15x-56=0[/tex]
so
[tex]a=-1\\b=-15\\c=-56[/tex]
substitute in the formula
[tex]x=\frac{-(-15)\pm\sqrt{-15^{2}-4(-1)(-56)}} {2(-1)}[/tex]
[tex]x=\frac{15\pm\sqrt{1}} {-2}[/tex]
[tex]x=\frac{15\pm1} {-2}[/tex]
[tex]x=\frac{15+1} {-2}=-8[/tex]
[tex]x=\frac{15-1} {-2}=-7[/tex]
The roots are
x=-8 and x=-7
so
The quadratic equation in factored form is equal to
[tex]-x^2-15x-56=(x+8)(x+7)[/tex]