Question

1. factor each of the following, if possible. check with distribution.

a) $y^{2}+4y - 21$
b) $y^{2}+ty + 21$
c) $2w^{2}-8w + 42$
d) $x^{2}+3xy - 10y^{2}$

1. create your own example of a factoring problem similar to the ones in question (7) above. then factor it.

Explanation:

Step1: Factor a)

Find two numbers that multiply to - 21 and add to 4. The numbers are 7 and - 3. So $y^{2}+4y - 21=y^{2}+7y-3y - 21=y(y + 7)-3(y + 7)=(y + 7)(y - 3)$

Step2: Analyze b)

For the quadratic $y^{2}+ty + 21$, assume it factors as $(y + m)(y + n)=y^{2}+(m + n)y+mn$. We need $mn = 21$. The factor - pairs of 21 are (1,21) and (3,7). But there are no two integers $m$ and $n$ such that $m + n=t$ (in general non - integer $t$ cases), and over the integers, it cannot be factored for most values of $t$.

Step3: Factor c)

First, factor out the greatest common factor 2 from $2w^{2}-8w + 42$ to get $2(w^{2}-4w + 21)$. Then, for $w^{2}-4w + 21$, find two numbers that multiply to 21 and add to - 4. The factor - pairs of 21 are (1,21) and (3,7). We rewrite $w^{2}-4w + 21$ as $w^{2}-3w-7w + 21=w(w - 3)-7(w - 3)=(w - 3)(w - 7)$. So $2w^{2}-8w + 42=2(w - 3)(w - 7)$

Step4: Factor d)

For $x^{2}+3xy-10y^{2}$, consider it as a quadratic in $x$ with $a = 1$, $b = 3y$, $c=-10y^{2}$. We need two numbers that multiply to $-10y^{2}$ and add to $3y$. The numbers are $5y$ and $-2y$. So $x^{2}+3xy-10y^{2}=x^{2}+5xy-2xy - 10y^{2}=x(x + 5y)-2y(x + 5y)=(x + 5y)(x - 2y)$

Answer:

a) $(y + 7)(y - 3)$
b) Cannot be factored over the integers
c) $2(w - 3)(w - 7)$
d) $(x + 5y)(x - 2y)$