ACT Math 3.4 Elementary Algebra
ACT Math Elementary Algebra Drill 3, Problem 4. Solve for y in this equation.
|ACT Math||Elementary Algebra|
|ACT Mathematics||Elementary Algebra|
|Algebra||Understand the relationship between zeros and factors of polynomials|
Use polynomial identities to solve problem
Use polynomial identities to solve problems
|Elementary Algebra||Solve quadratic equations by factoring|
Solving simple quadratic equations
|Foreign Language||Arabic Subtitled|
|Product Type||ACT Math|
We just have to figure out what the factors are... and there's a bit of a curveball in
here in that there's a 2 on the front end.
So we know that one of the factors is going to be a 2, as in:
The quantity 2y plus something times the quantity y plus something.
Let's plug in some numbers to get the 3. It's kind of easy because 3 only has 2 factors:
1 and 3...
Now we have quantity 2y plus 1 times quantity y plus 3.
If we multiply it out, we get 2y times y here, which is 2y squared.
Then we have 2y times 3, which is 6y
plus y times 1, which is just... y.
6y plus y is 7y; then, 1 times 3 is just 3 so we're done.
But now... how do we make the whole thing equal to zero? Well, one of the elements here
has to BE zero.
If y is negative 3 then this one is 3 minus 3 which is 0...
...and here, it's just negative one half, because negative 1 half times 2 is negative
1... and negative 1 plus positive 1 is zero.
So the other answer is negative one half and the answer is A and C...
...or option E.