# how to factor

• Nov 20th 2006, 09:23 PM
shenton
how to factor
How do you factor these? The split middle term doesn't work.

Q1:
-6a^2 - 5a -1
product = 6, sum = -5

-6a^2 + a - 6a -1
(-6a^2 + a) - (6a + 1)
a(-6a + 1) - (6a + 1)
(-6a + 1) (a - 1) -> this is not giving me the same answer if foil

Q2:
25a^2 -10a + 1
product = 25, sum = -10
?? -> cannot find 2 numbers whose product is 25 and difference is -10

Q3:
12x^2 +17xy + 6y^2
?? -> Two terms with ^2, how do you do this?

Q4:
-30x^3 + 2x^2 + 12x
?? -> One term with ^3 and another with ^2, how do you do this?

Thanks.
• Nov 21st 2006, 02:56 AM
galactus
Quote:

Originally Posted by shenton
How do you factor these? The split middle term doesn't work.

Q1:
-6a^2 - 5a -1
product = 6, sum = -5

-6a^2 + a - 6a -1
(-6a^2 + a) - (6a + 1)
a(-6a + 1) - (6a + 1)
(-6a + 1) (a - 1) -> this is not giving me the same answer if foil

Ask yourself, what two numbers when added equal -5 and when multiplied equal 6(the leading coefficient of -6 must be accounted for, -6(-1)=6).

Let's see...-3 and -2.....(-3)(-2)=6 and -3+(-2)=-5

$\displaystyle -6x^{2}-3x-2x-1$

$\displaystyle (-6x^{2}-3x)-(2x+1)$

Factor:

$\displaystyle -3x(2x+1)-(2x+1)$

$\displaystyle (-3x-1)(2x+1)$

Quote:

Q4:
-30x^3 + 2x^2 + 12x
?? -> One term with ^3 and another with ^2, how do you do this?
First, factor out 2x:

$\displaystyle 2x(-15x^{2}+x+6)$

Now, the quadratic in the parentheses factors:

Same as before:

What 2 numbers when multiplied equal -90,(-15*6), and when added equal 1.

10 and -9?. Yep.

$\displaystyle -15x^{2}+10x-9x+6$

Factor:

$\displaystyle 5x(-3x+2)+3(-3x+2)$

$\displaystyle (5x+3)(-3x+2)$

$\displaystyle 2x(5x+3)(-3x+2)$
• Nov 21st 2006, 03:02 AM
malaygoel
Quote:

Originally Posted by shenton
How do you factor these? The split middle term doesn't work.

Q1:
-6a^2 - 5a -1
product = 6, sum = -5

-6a^2 + a - 6a -1
(-6a^2 + a) - (6a + 1)
a(-6a + 1) - (6a + 1)
(-6a + 1) (a - 1) -> this is not giving me the same answer if foil

the answer you got is right.
the other form is (6a-1)(1-a)
And what do you mean by "if foil"..............sorry for the mistake
Quote:

Q2:
25a^2 -10a + 1
product = 25, sum = -10
?? -> cannot find 2 numbers whose product is 25 and difference is -10
difference is not -10 but the sum is -10. Now, i think you can easily find the numbers.
Quote:

Q3:
12x^2 +17xy + 6y^2
?? -> Two terms with ^2, how do you do this?
You apply the same method as in above problems.
the general expression is
$\displaystyle ax^2 + bxy + cy^2$
you factorise it by finding the two numbers such h and k such that
h+k=b
hk=ac
or by spliiting the terms(xy term is split)
and the factors are of the form: (ex+fy)(gx+hy)
the problems you were doing have y=1[/quote]
Quote:

Q4:
-30x^3 + 2x^2 + 12x
?? -> One term with ^3 and another with ^2, how do you do this?
you can take x common in every term to get
x(-30x^2+2x+12)
now you can simply factorise

Keep Smiling
Malay
• Nov 21st 2006, 04:07 AM
topsquark
Quote:

Originally Posted by shenton
Q1:
-6a^2 - 5a -1
product = 6, sum = -5

-6a^2 + a - 6a -1
(-6a^2 + a) - (6a + 1)
a(-6a + 1) - (6a + 1)
(-6a + 1) (a - 1) -> this is not giving me the same answer if foil

Take a look at the line in red. You can't factor the two because they aren't the same. You effectively stated that -(6a + 1) = -6a + 1, which is not correct. galactus' solution is right.

Quote:

Originally Posted by shenton
Q2:
25a^2 -10a + 1
product = 25, sum = -10
?? -> cannot find 2 numbers whose product is 25 and difference is -10

$\displaystyle (-5) \cdot (-5) = 25$ and $\displaystyle (-5) + (-5) = -10$

So
$\displaystyle 25a^2 -10a + 1 = 25a^2 - 5a - 5a + 1$

= $\displaystyle 5a(5a - 1) - (5a - 1) = (5a - 1)(5a - 1) = (5a - 1)^2$

-Dan
• Nov 21st 2006, 11:18 AM
shenton
Thanks guys for the guidance.

Thanks galactus for showing Q1 and Q4. Question 1 is a big help because I kept using 1 and 6 instead of 2 and 3.

Thanks malaygoel for your help.

Quote:

And what do you mean by "if foil"
FOIL of (a + b)(a + b) means multiplying the factors out, ie.
(a + b)(a + b) = a^2 +ab + ba + b^2. We should get back the original and this would enable us to verify that our factors are right.

Quote:

you can take x common in every term to get x(-30x^2+2x+12)
galactus has a better way of doing this. He took 2x instead of x. This gives us (15x^2 + x + 6). I guess we need to take the the "biggest" or all the common terms such as 2x is more than x.

Thanks topsquark as usual for the help.

Quote:

a(-6a + 1) - (6a + 1) Take a look at the line in red. You can't factor the two because they aren't the same.
This comment is great help. I've always wondered and now I know that during the process of factoring, the two ( ) ( ) must be the same. If it is not the same, somewhere is wrong and they are not factors.

Thanks for showing Q2.

Quote:

Q3:
12x^2 +17xy + 6y^2
?? -> Two terms with ^2, how do you do this?
Nobody worked on Q2. This question I belive is different from the rest.

12x^2 +17xy + 6y^2

We have a x^2 and y^2.

It does not fall under ax^2 + bx + c style where I work by getting product and sum.

How do you do this in the proper method (other than trial and error)?

Thanks.
• Nov 21st 2006, 11:49 AM
topsquark
Quote:

Originally Posted by shenton
12x^2 +17xy + 6y^2

We have a x^2 and y^2.

It does not fall under ax^2 + bx + c style where I work by getting product and sum.

How do you do this in the proper method (other than trial and error)?

Thanks.

Actually it IS quite similar (if you've done one before.)

Note that the factorization has to take the form of:
$\displaystyle (ax + by)(cx + dy)$
else we aren't going to get the correct kind of terms. (We might get terms like $\displaystyle 7x^2y$ for example.)

So work your factoring like you did before:
You are looking for two numbers that have the product: $\displaystyle 12 \cdot 6 = 72$ and sum to 17.

This is a little hairier than the other examples, so I think the simplest thing to do would be to start listing the factors of 72 in pairs:
1, 72 -> 1 + 72 = 73
2, 36 -> 2 + 36 = 38
etc.

I find that the pair 8,9 has a product of 72 and sums to 17.

So:
$\displaystyle 12x^2 +17xy + 6y^2 = 12x^2 + 8xy + 9xy + 6y^2$

= $\displaystyle (12x^2 + 8xy) + (9xy + 6y^2)$

= $\displaystyle 4x(3x + 2y) + 3y(3x + 2y)$

= $\displaystyle (4x + 3y)(3x + 2y)$

You can multiply this out to check that it gives the correct answer.

-Dan
• Nov 21st 2006, 12:09 PM
shenton
Thanks, topsquark again for showing how to deal with this type of question.

Quote:

This is a little hairier than the other examples
This type of question is certainly hairy.. it strecthes one's imagination to find the factors. At least for me. I spend much more time working on these questions than other topics (exponents, radicals). And now I have a bunch of such questions to work on. Thanks for the rescue.
• Nov 21st 2006, 12:57 PM
topsquark
Quote:

Originally Posted by shenton
Thanks, topsquark again for showing how to deal with this type of question.

This type of question is certainly hairy.. it strecthes one's imagination to find the factors. At least for me. I spend much more time working on these questions than other topics (exponents, radicals). And now I have a bunch of such questions to work on. Thanks for the rescue.

Some of this stuff is analytical and some of it falls under the category "you have to see it done once before you can tackle it." Just hang in there and keep asking questions and you'll do fine. :)

-Dan
• Nov 21st 2006, 01:42 PM
Quick
Factoring Tip!

I figured this out when I wasn't paying attention in math :p

if you want to split $\displaystyle z_1x^2+z_2x+z_3$ into $\displaystyle (ax+b)(cx+d)$

You can split it into $\displaystyle z_1x^2+\left(\frac{z_2+\sqrt{z_2^2-4z_1z_3}}{2}\right)x+\left(\frac{z_2-\sqrt{z_2^2-4z_1z_3}}{2}\right)x+z_3$

note that either $\displaystyle \frac{z_2+\sqrt{z_2^2-4z_1z_3}}{2}=ad$ or $\displaystyle \frac{z_2+\sqrt{z_2^2-4z_1z_3}}{2}=bc$

let's say we have: $\displaystyle -6a^2 - 5a -1$

So you can split it into this: $\displaystyle z_1x^2+\left(\frac{z_2+\sqrt{z_2^2-4z_1z_3}}{2}\right)x+\left(\frac{z_2-\sqrt{z_2^2-4z_1z_3}}{2}\right)x+z_3$

substitute: $\displaystyle -6x^2+\left(\frac{-5+\sqrt{(-5)^2-4(-6)(-1)}}{2}\right)x+$$\displaystyle \left(\frac{-5-\sqrt{(-5)^2-4(-6)(-1)}}{2}\right)x-1 thus: \displaystyle -6x^2+\left(\frac{-5+\sqrt{25-24}}{2}\right)x+\left(\frac{-5-\sqrt{25-24}}{2}\right)x-1 therefore: \displaystyle -6x^2+\left(\frac{-5+\sqrt{1}}{2}\right)x+\left(\frac{-5-\sqrt{1}}{2}\right)x-1 thus: \displaystyle -6x^2+\left(\frac{-5+1}{2}\right)x+\left(\frac{-5-1}{2}\right)x-1 then: \displaystyle -6x^2+\left(\frac{-4}{2}\right)x+\left(\frac{-6}{2}\right)x-1 then: \displaystyle -6x^2-2x-3x-1 note that it is now in either the form: \displaystyle acx^2+adx+bcx+bd or \displaystyle acx^2+bcx+adx+bd • Nov 21st 2006, 02:16 PM topsquark Quote: Originally Posted by Quick Factoring Tip! I figured this out when I wasn't paying attention in math :p if you want to split \displaystyle z_1x^2+z_2x+z_3 into \displaystyle (ax+b)(cx+d) You can split it into \displaystyle z_1x^2+\left(\frac{z_2+\sqrt{z_2^2-4z_1z_3}}{2}\right)x+\left(\frac{z_2-\sqrt{z_2^2-4z_1z_3}}{2}\right)x+z_3 note that either \displaystyle \frac{z_2+\sqrt{z_2^2-4z_1z_3}}{2}=ad or \displaystyle \frac{z_2+\sqrt{z_2^2-4z_1z_3}}{2}=bc let's say we have: \displaystyle -6a^2 - 5a -1 So you can split it into this: \displaystyle z_1x^2+\left(\frac{z_2+\sqrt{z_2^2-4z_1z_3}}{2}\right)x+\left(\frac{z_2-\sqrt{z_2^2-4z_1z_3}}{2}\right)x+z_3 substitute: \displaystyle -6x^2+\left(\frac{-5+\sqrt{(-5)^2-4(-6)(-1)}}{2}\right)x+$$\displaystyle \left(\frac{-5-\sqrt{(-5)^2-4(-6)(-1)}}{2}\right)x-1$

thus: $\displaystyle -6x^2+\left(\frac{-5+\sqrt{25-24}}{2}\right)x+\left(\frac{-5-\sqrt{25-24}}{2}\right)x-1$

therefore: $\displaystyle -6x^2+\left(\frac{-5+\sqrt{1}}{2}\right)x+\left(\frac{-5-\sqrt{1}}{2}\right)x-1$

thus: $\displaystyle -6x^2+\left(\frac{-5+1}{2}\right)x+\left(\frac{-5-1}{2}\right)x-1$

then: $\displaystyle -6x^2+\left(\frac{-4}{2}\right)x+\left(\frac{-6}{2}\right)x-1$

then: $\displaystyle -6x^2-2x-3x-1$

note that it is now in either the form: $\displaystyle acx^2+adx+bcx+bd$ or $\displaystyle acx^2+bcx+adx+bd$

You realize, of course, that this works because if you have $\displaystyle (x - r_1)(x - r_2) = x^2 - (r_1 + r_2)x + r_1r_2$

The negative sum of the zeros will be the middle coefficient.

However, if you insist on doing it this way I must point out it is far simpler just to solve for the zeros using the quadratic formula.

-Dan
• Nov 21st 2006, 02:45 PM
Quick
Quote:

Originally Posted by topsquark
You realize, of course, that this works because if you have $\displaystyle (x - r_1)(x - r_2) = x^2 - (r_1 + r_2)x + r_1r_2$

The negative sum of the zeros will be the middle coefficient.

However, if you insist on doing it this way I must point out it is far simpler just to solve for the zeros using the quadratic formula.

-Dan

my way works for (ax+b)(cx+d) and (ax-b)(cx-d) and (ax+b)(cx-d)

not just (ax-b)(cx-d)

but I admit it's slow :o it's meant to help you if your stuck.