Factoring polynomials

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• Jan 13th 2008, 09:20 PM
MathMack
Factoring polynomials
Hey, I have a question, you can factor using guess and check or grouping, here it is

-13a + 12 - 4a^2

the answer is -(a+4)(4a-3)

Not sure how to get it using either one, please help! Thanks!
• Jan 13th 2008, 10:02 PM
janvdl
Quote:

Originally Posted by MathMack
Hey, I have a question, you can factor using guess and check or grouping, here it is

\$\displaystyle -13a + 12 - 4a^2\$

the answer is -(a+4)(4a-3)

Not sure how to get it using either one, please help! Thanks!

First re-arrange:

\$\displaystyle -4a^2 - 13a + 12 \$

Now factor out the negative sign

\$\displaystyle -(4a^2 + 13a - 12) \$

Now we should take the \$\displaystyle 4a^2 \$ term and factorise that, keeping in mind that it is not yet complete. (The \$\displaystyle x_{1} \$ and \$\displaystyle x_{2} \$ terms represent the constants)

\$\displaystyle -(4a + x_{1})(a + x_{2}) \$

Now we should see which constants give us a product of 12.
(1 x 12 ; 2 x 6 ; 3 x 4)

Now from those options we should find which also satisfies the middle term.

Can you take it from here? Just do it step by step.
• Jan 13th 2008, 10:03 PM
MathMack
I also have 4 more, any help is appreciated!

2.) n^3-n^2+n-1

3.) 4a(a+2) - 8 (a+2)

4.) -6x^3 +10x^2 - 4x^4

5.) a rectangle has area described by the polynomial 6x^2 + x -1 suare centimeters. Find the length and width of the rectangle. Thanks again
• Jan 13th 2008, 10:08 PM
MathMack
kinda still stuck on that first one. Thanks for your help
• Jan 13th 2008, 10:11 PM
janvdl
Quote:

Originally Posted by MathMack
I also have 4 more, any help is appreciated!

2.) n^3-n^2+n-1

3.) 4a(a+2) - 8 (a+2)

4.) -6x^3 +10x^2 - 4x^4

5.) a rectangle has area described by the polynomial 6x^2 + x -1 suare centimeters. Find the length and width of the rectangle. Thanks again

2)\$\displaystyle n^2(n - 1) + 1(n - 1)\$
\$\displaystyle = (n^2 + 1)(n - 1)\$

3) \$\displaystyle (4a - 8)(a + 2)\$

4) \$\displaystyle -2(3x^3 - 5x^2 + 2x^4)\$
\$\displaystyle = -2x^2(3x - 5 + 2x^2)\$
\$\displaystyle = -2x(2x^2 + 3x - 5)\$
\$\displaystyle = -2x(2x + 5)(x - 1)\$

5) \$\displaystyle 6x^2 + x - 1\$

We know that this value is equal to 2 values multiplied by each other. So let's factorise it:

\$\displaystyle (3x - 1)(2x + 1)\$
• Jan 13th 2008, 10:12 PM
janvdl
Quote:

Originally Posted by MathMack
kinda still stuck on that first one. Thanks for your help

Which part do you not understand? Give me the exact part where you got stuck and why you do not understand it.
• Jan 13th 2008, 10:15 PM
MathMack
I got the part where you take the negative out, but after that i'm lost, I know it's (4a something)(a something) but lost to go after that, I know you must find factors but not sure
• Jan 13th 2008, 10:24 PM
janvdl
Quote:

Originally Posted by MathMack
I got the part where you take the negative out, but after that i'm lost, I know it's (4a something)(a something) but lost to go after that, I know you must find factors but not sure

You must find the factors of the 12 at the end. Then from all those factors, you must find the pair which satisfies the middle term too.

In other words: When you expand the brackets again, you should end up with the same expression that you started with.
• Jan 13th 2008, 10:25 PM
MathMack
and the answers in the back of the book for the problems you answered are a little different, thanks for helping me, i'm just confused!

2.) n^3-n^2+n-1
(n-1) (n^2 +1)
3.) 4a(a+2) - 8 (a+2)
4(a+2)(a-2)
4.) -6x^3 +10x^2 - 4x^4
-2x^2(2x+5)(x-1)
5.) a rectangle has area described by the polynomial 6x^2 + x -1 suare centimeters. Find the length and width of the rectangle.
2x-5
• Jan 13th 2008, 10:27 PM
janvdl
Quote:

Originally Posted by MathMack
and the answers in the back of the book for the problems you answered are a little different, thanks for helping me, i'm just confused!

2.) n^3-n^2+n-1
(n-1) (n^2 +1)
3.) 4a(a+2) - 8 (a+2)
4(a+2)(a-2)
4.) -6x^3 +10x^2 - 4x^4
-2x^2(2x+5)(x-1)
5.) a rectangle has area described by the polynomial 6x^2 + x -1 suare centimeters. Find the length and width of the rectangle.
2x-5

3) In number 3 they merely factorised (4a - 8) to 4(a - 2)

5) Is the answer (2x - 5) ??
• Jan 13th 2008, 10:30 PM
MathMack
yes and on the first one we worked on they have the answer -(a+4) (4a-3)

Still lost on that one
THANKS JANVDLL!
• Jan 13th 2008, 10:39 PM
janvdl
Quote:

Originally Posted by MathMack
yes and on the first one we worked on they have the answer -(a+4) (4a-3)

Still lost on that one
THANKS JANVDLL!

You're welcome but I'm more interested in making you understand.

Tell me why 4 and -3 are the constants in the brackets?
• Jan 13th 2008, 10:41 PM
MathMack
that's what i'm not sure on either, the part that we should find two numbers for is positive 12
• Jan 13th 2008, 10:45 PM
janvdl
Quote:

Originally Posted by MathMack
that's what i'm not sure on either, the part that we should find two numbers for is positive 12

But remember we factored a negative out...

So then \$\displaystyle -(4a^2 + 13a - 12)\$

Now just put that negative sign aside(But don't throw it away!).

Imagine the equation would be only \$\displaystyle 4a^2 + 13a - 12 \$

Now can you see why we use 4 and -3 ?
• Jan 13th 2008, 10:54 PM
MathMack
okay, I finally got that one all figured out completly, just took awhile, The next one I can't figure out is the n^3-n^2 +n -1
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