Question

solving by factoring (a = 1)$m^{2}-7m-18=0$$f(x)=x^{2}-5x-14$$f(p)=p^{2}-9p+8$$0=z^{2}-8z+16$

Explanation:

Step1: Factor $m^2-7m-18=0$

Find two numbers that multiply to $-18$ and add to $-7$: $-9$ and $2$.
$\implies (m-9)(m+2)=0$
Set each factor to 0:
$m-9=0 \implies m=9$; $m+2=0 \implies m=-2$

Step2: Factor $f(x)=x^2-5x-14$

Find two numbers that multiply to $-14$ and add to $-5$: $-7$ and $2$.
$\implies f(x)=(x-7)(x+2)$
To find roots, set $f(x)=0$:
$x-7=0 \implies x=7$; $x+2=0 \implies x=-2$

Step3: Factor $f(p)=p^2-9p+8$

Find two numbers that multiply to $8$ and add to $-9$: $-8$ and $-1$.
$\implies f(p)=(p-8)(p-1)$
To find roots, set $f(p)=0$:
$p-8=0 \implies p=8$; $p-1=0 \implies p=1$

Step4: Factor $0=z^2-8z+16$

This is a perfect square trinomial: $(-4)^2=16$ and $2(-4)=-8$.
$\implies (z-4)^2=0$
Set factor to 0:
$z-4=0 \implies z=4$ (repeated root)

Answer:

1. For $m^2-7m-18=0$: $m=9$ or $m=-2$
2. For $f(x)=x^2-5x-14$: Factored form is $(x-7)(x+2)$; roots are $x=7$ or $x=-2$
3. For $f(p)=p^2-9p+8$: Factored form is $(p-8)(p-1)$; roots are $p=8$ or $p=1$
4. For $0=z^2-8z+16$: $z=4$ (double root)