factor the following\na) $x^{2}-5x - 6$\nb) $x^{2}-7x + 12$\nc) $x^{2}-20x + 100$\nd) $x^{2}-x - 56$\ne)…

Explanation:

Step1: Recall factoring formula for $x^2+bx + c$

For a quadratic expression $x^2+bx + c$, we need to find two numbers $m$ and $n$ such that $m + n=b$ and $m\times n = c$.

Step2: Factor $x^2-5x - 6$

We need two numbers $m$ and $n$ with $m + n=-5$ and $m\times n=-6$. The numbers are $-6$ and $1$. So $x^2-5x - 6=(x - 6)(x+1)$.

Step3: Factor $x^2-7x + 12$

We find two numbers $m$ and $n$ with $m + n=-7$ and $m\times n = 12$. The numbers are $-3$ and $-4$. So $x^2-7x + 12=(x - 3)(x - 4)$.

Step4: Factor $x^2-20x + 100$

This is a perfect - square trinomial. Since $x^2-20x + 100=x^2-2\times10x+10^2$, by the formula $(a - b)^2=a^2-2ab + b^2$, we have $x^2-20x + 100=(x - 10)^2$.

Step5: Factor $x^2-x - 56$

We need two numbers $m$ and $n$ with $m + n=-1$ and $m\times n=-56$. The numbers are $-8$ and $7$. So $x^2-x - 56=(x - 8)(x+7)$.

Step6: Factor $x^2-3x - 10$

We find two numbers $m$ and $n$ with $m + n=-3$ and $m\times n=-10$. The numbers are $-5$ and $2$. So $x^2-3x - 10=(x - 5)(x+2)$.

Step7: Factor $x^2+5x + 4$

We need two numbers $m$ and $n$ with $m + n=5$ and $m\times n = 4$. The numbers are $1$ and $4$. So $x^2+5x + 4=(x + 1)(x + 4)$.

Step8: Factor $x^2-11x + 24$

We find two numbers $m$ and $n$ with $m + n=-11$ and $m\times n = 24$. The numbers are $-3$ and $-8$. So $x^2-11x + 24=(x - 3)(x - 8)$.

Step9: Factor $x^2+10x - 11$

We need two numbers $m$ and $n$ with $m + n=10$ and $m\times n=-11$. The numbers are $11$ and $-1$. So $x^2+10x - 11=(x + 11)(x - 1)$.

Answer:

A. $(x - 6)(x+1)$ B. $(x - 3)(x - 4)$ C. $(x - 10)^2$ D. $(x - 8)(x+7)$ E. $(x - 5)(x+2)$ F. $(x + 1)(x + 4)$ G. $(x - 3)(x - 8)$ H. $(x + 11)(x - 1)$