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#1 2014-07-03 07:36:27

demha
Member
Registered: 2012-11-25
Posts: 192

Factoring Quadratic Equations

Hello everyone I'm back after a pretty long time big_smile

I haven't done math in quite a while so I'm going to need help with this. Basically I just need help to understand how to go through the equations step-by-step and then tell me if I'm right or wrong. Here they are:
12. 2x^2 + 23x - 12
A (2x - 6)(x + 2)
B(2x - 3)(x + 4) 
C(2x + 12)(x - 1) 
D(2x - 1)(x + 12)
E (x - 6)(x + 2)
F (2x - 1)(x - 12)

Answer:
2x^2 + 23x - 12
-1 and 24 are the common factors so:
2x^2 - x + 24x - 12
(2x^2 - x) + (24x - 12)
Factor Out:
x(2x - 1)+12(2x - 1)
The '(2x - 1)' is a common factor so we factor it out, being left with:
(2x - 1)(x + 12)
Final Answer Being D

Now from here on out I'm going to be needing some help.



13. a^2 + 4ab - 3b^2
A (a - 3b)(a - b)
B(a - b)(a + 3b)
C cannot be factored
D(a + b)(a + 3b)
E a(a + 4b) -3(-b)
F (a + b)(a - 3b)

I'm not sure how to get the answer using the equation problem. I do know how to work the factor answers backwards and came with the idea that the answer might be 'F.' I want to be sure if I am right and even if I am, I still want to understand how I can find the factors for this. I find it quite confusing mostly because of the double variables it has.


When we are done with the first two we can move on to the rest.

14. 9k^2 + 30kn + 25n^2
A (9k + 5n)(k + n)
B(3k - 5n)(3k + 5n)
C(3k - 5)(3k - 5)
D(3k + 5n)2 
E (9k + 25)(k + 1)
F (9k - 5)(9k - 5)


15. 2a^2-32
A 2(a^2 - 18)
B 2(a - 4)2
C (2a - 16)(a + 2)
D 2(a^2 - 16)
E 2(a - 4)(a + 4)
F cannot be factored


16. 6r^2 + 13r + 6
A 6(r^2 + 1) + 13
B(6r + 6)(r + 1)
C(2r + 3)(3r + 2)
D(2r + 2)(3r + 3)
E (6r + 1)(r + 6)
F cannot be factored


17. 4m^4 - p^2
A (4m2 - p)(m + p)
Bcannot be factored
C(2m^2 + p2)^2   
D(2m^2 - p2)^2
E (4m + p)(m3 - p)
F (2m^2-p)(2m^2+p)


18. 5n^2 + 19n - 4   
A (n + 4)(5n - 1)
B(5n - 4)(n + 1)
C(5n - 4)(n - 1)
D(n - 4)(5n + 1)   
E (n - 2)(5n + 2)
F cannot be factored


19. m^2 + (5/12)m - (1/6)    
A (m - 6)(m + 1)
B(m - 2/3)(m - 1/4)
Ccannot be factored
D(m - 1/4)(m + 2/3)
E (m/12 + 2)(m/12 - 3)
F (m - 1/4)(m - 1/3)


20. a^2 + 4a + 21
A (a + 7)(a - 3)
Ba(a + 4) + 21
C(a + 21)(a - 1)
D(a + 3)(a - 7)
E (a + 7)(a + 3)
F cannot be factored


"The thing about quotes on the Internet is you cannot confirm their validity"
~Abraham Lincoln

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#2 2014-07-04 02:14:07

Maburo
Member
From: Alberta, Canada
Registered: 2013-01-08
Posts: 287

Re: Factoring Quadratic Equations

Hello demha,

Don't let the double variables confuse you. Double variable quadratics can still be factored the same way.

13.   We need to look for two numbers which multiply to produce

and sum to
. No such real numbers exist, so the answer is C - cannot be factored.


14.   This should be a better example to rid any confusion surrounding double variable quadratics, since this one can be factored. Again, we must look for two numbers which multiply to produce

and sum to
. Two such numbers are
and
because
and
. Once we have found these numbers, we replace the middle term with our new numbers as coefficients:

Then factor terms with k and n:

Now factor out the common factor of
:

And the answer is then D. See, it's no different with two variables!


15.   This one is a little different in that it uses the concept of "difference of squares." We have


We factor out a 2:

The difference of squares basically states that
. In this case,
, so we can write

And thus the answer is E.


16.   I'm sure you can do this one! Try it out.


17.   Remember the difference of squares? We need it to factor this one! We have

. To use the difference of squares we need
. This is possible if we let
and
. Then

or

Thus, the answer is F.


Try doing the rest on your own. I don't see any more with double variables, but if you ever run into them, you know they shouldn't be treated any differently!


"Pure mathematics is, in its way, the poetry of logical ideas."
-Albert Einstein

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#3 2014-07-04 04:17:12

demha
Member
Registered: 2012-11-25
Posts: 192

Re: Factoring Quadratic Equations

Awesome! Thank you very much for the help! It's definitely much easier for me now.

For 16, I got the answer 'C' [(2r + 3)(3r + 2) ].

For 18:
18. 5n^2 + 19n - 4   
A (n + 4)(5n - 1)
B (5n - 4)(n + 1)
C (5n - 4)(n - 1)
D (n - 4)(5n + 1)   
E (n - 2)(5n + 2)
F cannot be factored

I choose 'F' because there is nothing I can think of that will add up to 4 when times together and add up to 19 when added together.

19:
19. m^2 + (5/12)m - (1/6)    
A (m - 6)(m + 1)
B (m - 2/3)(m - 1/4)
C cannot be factored
D (m - 1/4)(m + 2/3)
E (m/12 + 2)(m/12 - 3)
F (m - 1/4)(m - 1/3)

I was having trouble trying to find the factors for these, I'm not too great with fractions.


20:
20. a^2 + 4a + 21
A (a + 7)(a - 3)
B a(a + 4) + 21
C (a + 21)(a - 1)
D (a + 3)(a - 7)
E (a + 7)(a + 3)
F cannot be factored   


Answer is 'F.'


"The thing about quotes on the Internet is you cannot confirm their validity"
~Abraham Lincoln

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#4 2014-07-04 05:13:12

Maburo
Member
From: Alberta, Canada
Registered: 2013-01-08
Posts: 287

Re: Factoring Quadratic Equations

For 16, C is correct. Great!


For 18, you must also consider the coefficient for

. You need two numbers which multiply to produce
and sum to the middle coefficient for
,
. We can, in fact, find these two numbers! They are
and
because
and
. Then we replace the coefficient of
with the new numbers we found, and we get:

Then factor out
:

Then factor out
:

Then the answer is A


19.   Yes, fractions can be quite tricky. Plus, the hardest part of factoring these quadratics is finding the new coefficients for the middle term. So mix that with fractions, and they can be a little nasty. Anyhow, let's follow the same steps and see what we can come up with. The coefficient in front of

is
and the constant is
, so we need two numbers that multiply to produce
and sum to the middle coefficient,
. Two numbers which have these properties are
and
because
, and
. Again, we then replace the coefficient of the middle term with our new numbers:

Then factor out
from the first two terms, and
from the last two:

Then factor out
:

Thus, the answer is D.


20.   You are correct!


"Pure mathematics is, in its way, the poetry of logical ideas."
-Albert Einstein

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#5 2014-07-04 06:42:37

demha
Member
Registered: 2012-11-25
Posts: 192

Re: Factoring Quadratic Equations

Yeah the fraction was really confusing for me. Thank you very much for helping me out on this Maburo!


"The thing about quotes on the Internet is you cannot confirm their validity"
~Abraham Lincoln

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#6 2014-07-04 07:04:51

Maburo
Member
From: Alberta, Canada
Registered: 2013-01-08
Posts: 287

Re: Factoring Quadratic Equations

No problem, demha!


"Pure mathematics is, in its way, the poetry of logical ideas."
-Albert Einstein

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