Algebra I Question

[r4dioh34d]

I am having trouble on a question on my Algebra I review.
#51. Open Ended... Find three different values to complete the expression x^2+?x+30 so that it can be factored into the product of two binomials. Show each factorization.

I have no idea how to solve for "?"

Can someone give me a hand?

I appreciate any help, thanks.

 

[pappus]

I am having trouble on a question on my Algebra I review.
#51. Open Ended... Find three different values to complete the expression x^2+?x+30 so that it can be factored into the product of two binomials. Show each factorization.

I have no idea how to solve for "?"

Can someone give me a hand?

I appreciate any help, thanks.

1. You are looking for a term in the form

\(\displaystyle (x+a)(x+b)= x^2+(a+b)x + ab\)

2. You know that

\(\displaystyle a \cdot b = 30\)

Choose a and determine b:

\(\displaystyle \begin{array}{c|cccc}a&1&2&3&5&... \\ b&30&15&10&6&... \end{array}\)

3. Now evaluate the term in the brackets. For instance: a = 1, b = 30. You'll get:

\(\displaystyle x^2+(1+30)x+30 = (x+1)(x+30)\)

 

[soroban]

Hello, r4dioh34d!

#51. Find three different values to complete the expression \(\displaystyle x^2+?x+30\)
so that it can be factored into the product of two binomials. .Show each factorization.

If you were given the middle coefficient, could you factor the trinomial?

Example:. . \(\displaystyle x^2 + 11x + 30\)

You would factor 30 into two parts whose sum is 11 . . . right?
You would get 5 and 6.

The factors are: .\(\displaystyle (x+5)(x+6)\)


In this problem, the middle coefficient is not given.
So, factor 30 into two parts . . . any two parts!


\(\displaystyle \begin{array}{cccccccc}30 \:=\5)(6) &\Rightarrow& (x+5)(x+6) &=& x^2 + 11x + 30 \\ \\ 30 =(\text{-}5)(\text{-}6) &=& (x-5)(x-6) &=& x^2-11x+30 \\ \\ 30 \:=\3)(10) &\Rightarrow& (x+3)(x+10) &=& x^2+13x+30 \\ \\ 30=(\text{-}3)(\text{-}10) &\Rightarrow& (x-3)(x-10) &=& x^2 - 13x + 30 \\ \\ 30\:=\1)(30) &\Rightarrow& (x+1)(x+30) &=& x^2 + 31x + 30 \\ \\ 30=(\text{-}1)(\text{-}30) &\Rightarrow& (x-1)(x-30) &=&x^2-31x + 30 \end{array}\)