Respuesta :

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]

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