We have two expressions.
The first expression is:
[tex]y^2-2y-8[/tex]Let's start with this first expression and review the process to factorize it, and then we will also factorize the second expression.
The process to factor an expression such as the two we have in this problem:
• Open two pairs of parentheses.
,• Put "y" at the beginning of each parenthesis
,• Find two numbers such that when you multiply them you get the independent number: -8. And when you add them you get the coefficient that accompanies the y: -2.
,• Add those numbers to the parentheses we previously opened.
For the first expression
[tex]y^2-2y-8[/tex]The first step is to open two parentheses and put y on each of them:
[tex]y^2-2y-8=(y\text{ }\ldots\text{)(y}\ldots\text{)}[/tex]Then we need to find two numbers that when you multiplied them, the result is -8, and when you add them the result is -2. Those numbers are -4 and +2, because:
-4x2=-8
-4+2=-2
And we complete the factored expression by adding these numbers to the parentheses:
[tex]y^2-2y-8=(y-4)(y+2)[/tex]The second expression is:
[tex]y^2+3y+2[/tex]We follow the same procedure to factor this expression.
First, open two pairs of parentheses and put y on each of them:
[tex]y^2+3y+2=(y\ldots)(y\ldots)[/tex]Then, look for two numbers that when you multiply them, the result is 2, and when you add them, the result is 3.
Those numbers are 2 and 1, because:
2x1=2
2+1=3
so we add them to our factored expression:
[tex]y^2+3y+2=(y+2)(y+1)[/tex]If you need to corroborate that this result is correct, you can multiply the factors on the right and check that you get the result on the left side.
Multiplying the factors to corroborate the answer:
[tex](y+2)(y+1)[/tex]We use the distributive property to multiply each term on the first parentheses by each term on the second parentheses:
[tex](y+2)(y+1)=y\cdot y+y\cdot1+2\cdot y+2\cdot1[/tex]and solving the multiplications:
[tex](y+2)(y+1)=y^2+y+2y+2[/tex]Combining the like terms:
[tex](y+2)(y+1)=y^2+3y+2[/tex]