3y^2+31y+36
Solve for y over the real numbers:
3 y^2+31 y+36 = 0
The left hand side factors into a product with two terms:
(y+9) (3 y+4) = 0
Split into two equations:
y+9 = 0 or 3 y+4 = 0
Subtract 9 from both sides:
y = -9 or 3 y+4 = 0
Subtract 4 from both sides:
y = -9 or 3 y = -4
Divide both sides by 3:
Answer: |
| y = -9 or y = -4/3
+118723 3y^2+31y+36
I am sorry guest but your answer is NOT correct.
There is no equal sign here, this is NOT an equation. So you have answered a different question entirely!
The question is, how do you factor \(3y^2+31y+36\).
What you need to look for is two numbers that multiply to 3*36 and add to 31
Since both these numbersare positive, both the numbers that you are looking for will also be positive.
Factors of 3*36
=3*3*12
=3*3*3*4
=3*3*3*2*2
= 27 * 4
and 27+4=31 so these are the numbers that I want.
So now I have to repleace 31y with 27y+4 like this
\(\quad3y^2+31y+36\\ =3y^2+27y+4y+36\\ \mbox{Now I have to factor in pairs}\\ =3y(y+9)+4(y+9)\\ =(3y+4)(y+9)\)
If you do not understand say so :))
