Answer:
A) x = -9, x = 8
B) x = 2, x = 10
C) x = -6, x = 10
D) x = -8, x = -3
E) x = -8, x = 9
Step-by-step explanation:
[tex]\boxed{\begin{aligned}&\textsf{Given}: \quad &x^2+x-72& =0\\&\textsf{Rewrite the term in $x$}: \quad &x^2+9x-8x-72& =0\\&\textsf{Factor the first two and last two terms}: \quad &x(x+9)-8(x+9)& =0\\&\textsf{Factor out $(x+9)$}: \quad &(x-8)(x+9)& =0\\&\textsf{Apply the zero-product property}: \quad &(x-8) &=0 \implies x=8\\&&(x+9)&=0 \implies x=-9\\&\textsf{Therefore, the roots are}: \quad & x&=-9,\;8\end{aligned}}[/tex]
[tex]\boxed{\begin{aligned}&\textsf{Given}: \quad &x^2-12x+20&=0\\&\textsf{Rewrite the term in $x$}: \quad &x^2-10x-2x+20& =0\\&\textsf{Factor the first two and last two terms}: \quad &x(x-10)-2(x-10)& =0\\&\textsf{Factor out $(x-10)$}: \quad &(x-2)(x-10)& =0\\&\textsf{Apply the zero-product property}: \quad &(x-2) &=0 \implies x=2\\&&(x-10)&=0 \implies x=10\\&\textsf{Therefore, the roots are}: \quad & x&=2,\;10\end{aligned}}[/tex]
[tex]\boxed{\begin{aligned}&\textsf{Given}: \quad &x^2-4x-60&=0\\&\textsf{Rewrite the term in $x$}: \quad &x^2-10x+6x-60& =0\\&\textsf{Factor the first two and last two terms}: \quad &x(x-10)+6(x-10)& =0\\&\textsf{Factor out $(x-10)$}: \quad &(x+6)(x-10)& =0\\&\textsf{Apply the zero-product property}: \quad &(x+6) &=0 \implies x=-6\\&&(x-10)&=0 \implies x=10\\&\textsf{Therefore, the roots are}: \quad & x&=-6,\;10\end{aligned}}[/tex]
[tex]\boxed{\begin{aligned}&\textsf{Given}: \quad &x^2+24&=-11x\\&\textsf{Add $11x$ to both sides}: \quad &x^2+11x+24&=0\\&\textsf{Rewrite the term in $x$}: \quad &x^2+8x+3x+24& =0\\&\textsf{Factor the first two and last two terms}: \quad &x(x+8)+3(x+8)& =0\\&\textsf{Factor out $(x+8)$}: \quad &(x+3)(x+8)& =0\\&\textsf{Apply the zero-product property}: \quad &(x+3) &=0 \implies x=-3\\&&(x+8)&=0 \implies x=-8\\&\textsf{Therefore, the roots are}: \quad & x&=-8,\;-3\end{aligned}}[/tex]
[tex]\boxed{\begin{aligned}&\textsf{Given}: \quad &x^2-x-72& =0\\&\textsf{Rewrite the term in $x$}: \quad &x^2-9x+8x-72& =0\\&\textsf{Factor the first two and last two terms}: \quad &x(x-9)+8(x-9)& =0\\&\textsf{Factor out $(x-9)$}: \quad &(x+8)(x-9)& =0\\&\textsf{Apply the zero-product property}: \quad &(x+8) &=0 \implies x=-8\\&&(x-9)&=0 \implies x=9\\&\textsf{Therefore, the roots are}: \quad & x&=-8,\;9\end{aligned}}[/tex]