Learn How to Factor Quadratics: Step-by-Step Guide for Gr. 10 Math Homework

AI Thread Summary
The discussion focuses on factoring the quadratic equation 2x² + 5x - 12. Participants suggest various methods, including trial and error with binomials and using Vieta's formulas to find suitable factors. A specific technique called "slide and divide" is mentioned, where the leading coefficient is multiplied by the constant term to simplify the factoring process. The importance of recognizing that the question requires factoring rather than completing the square is emphasized. Overall, the conversation provides multiple strategies for tackling the problem effectively.
winston2020
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Homework Statement



Factor.
2x^{2}+5x-12

I just took a semester off from school and I feel dumb. My recommendation to anyone reading is don't do that.
Anyways, back to gr. 10 math :cry:

Normally if I was going to factor this I would try to eliminate the coefficient of x^{2} but it can't be divided evenly. IIRC you're supposed to take the factors of c and try to find 2 that add up to b, but that doesn't seem to be an option here either...

I just need a push in the right direction :)
 
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Take a look at the Binomial theorem. That should help you out.

Thanks
Matt
 
winston2020 said:

Homework Statement



Factor.
2x^{2}+5x-12

I just took a semester off from school and I feel dumb. My recommendation to anyone reading is don't do that.
Anyways, back to gr. 10 math :cry:

Normally if I was going to factor this I would try to eliminate the coefficient of x^{2} but it can't be divided evenly. IIRC you're supposed to take the factors of c and try to find 2 that add up to b, but that doesn't seem to be an option here either...

I just need a push in the right direction :)

This is fairly straightforward, since there aren't many possibilities for 2x2. Start with (2x + ?)(x + ?) and by trial and error, see if there is some way to factor -12 so that you get an x-term coefficient of + 5.
 
Start by:

2(x2+(5/2)x-6)=2[ (x+5/4)2-25/4-6 ]
 
njama said:
Start by:

2(x2+(5/2)x-6)=2[ (x+5/4)2-25/4-6 ]

The question asked him to factor the quadratic, not complete the square.
 
It is possible to factor quadratic trinomials whose lead coefficient is not 1 by multiplying the lead and tail coefficients together and finding a factor pair of that product that adds up to the middle coefficient.

Based on a set of equations by Viete:

Find u and v so that ax^2 + bx + c = ax^2 + ux + vx + c and

uv=ac and u+v=b.

Once u and v are found, then factor by grouping.

--Elucidus
 
My Teacher taught me a method called slide and divide.

2x^2 +5x -12, you multiply the coefficient of 2x^2 times -12.

X^2 +5x-24, from there you factor normally...Once you get the factors, you divide the number of by the coefficient you used before. One little trick that you must remember is if the number turns out to be not a whole number, you bring out the denominator and put in on the x. For example (x-3/2) would become (2x-3)
 
gabbagabbahey said:
The question asked him to factor the quadratic, not complete the square.
Yes, but that's a perfectly valid way to find the factors (and works even if the coefficients in the factors are not rational).
2((x+ 5/4)^2- 25/4- 6)= 2((x+ 5/4)^2- 49/4) and that is now a "difference of squares": 2(x+5/4+ 7/2)(x+5/4- 7/2)
 
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