Factoring and impossible equation!!!

**curlysue** · April 15th 2008, 10:19 AM

(20b(squared)-11b-3)
Sorry I don't know how to type the squared symbol...but thanks in advance!

**angel.white** · April 15th 2008, 10:26 AM

Originally Posted by curlysue

(20b(squared)-11b-3)
Sorry I don't know how to type the squared symbol...but thanks in advance!

$20b^2 -11b -3= (4b -3)(5b+1)$

**topsquark** · April 15th 2008, 10:26 AM

Originally Posted by curlysue

(20b(squared)-11b-3)
Sorry I don't know how to type the squared symbol...but thanks in advance!

$20b^2 - 11b - 3$

Multiply the coefficient of the leading term by the constant term:
20 * -3 = -60

List all the pairs of factors of -60:
1, -60
2, -30
3, -20
4, -15
5, -12
6, -10
10, -6
12, -5
15, -4
20, -3
30, -2
60, -1

Now we want to find the pair of factors in the list above that add up to be equal to the coefficient of the linear term, in this case -11. (If no such pair of factors is in the list then you cannot factor the quadratic.)
I've got 4 + -15 = -11.

So split the linear term of the quadratic into -11x = 4x - 15x.
$20b^2 - 11b - 3 = 20b^2 + 4b - 15b - 3$

Group the terms as follows:
$= (20b^2 + 4b) + (-15b - 3)$

Now factor from each group:
$= 4b(5b + 1) + (-3)(5b + 1)$

and note that both terms have a common factor, 5b + 1, so we can factor that from each term as well:
$= (4b - 3)(5b + 1)$

There you go!

-Dan

**angel.white** · April 15th 2008, 10:30 AM

Originally Posted by topsquark

$20b^2 - 11b - 3$

Multiply the coefficient of the leading term by the constant term:
20 * -3 = -60

List all the pairs of factors of -60:
1, -60
2, -30
3, -20
4, -15
5, -12
6, -10
10, -6
12, -5
15, -4
20, -3
30, -2
60, -1

Now we want to find the pair of factors in the list above that add up to be equal to the coefficient of the linear term, in this case -11. (If no such pair of factors is in the list then you cannot factor the quadratic.)
I've got 4 + -15 = -11.

So split the linear term of the quadratic into -11x = 4x - 15x.
$20b^2 - 11b - 3 = 20b^2 + 4b - 15b - 3$

Group the terms as follows:
$= (20b^2 + 4b) + (-15b - 3)$

Now factor from each group:
$= 4b(5b + 1) + (-3)(5b + 1)$

and note that both terms have a common factor, 5b + 1, so we can factor that from each term as well:
$= (4b - 3)(5b + 1)$

There you go!

-Dan

That's pretty cool, I actually didn't know about the ordered pair method.

**red_dog** · April 15th 2008, 10:31 AM

$20b^2-11b-3=20b^2+4b-15b-3=4b(5b+1)-3(5b+1)=(5b+1)(4b-3)$

**wingless** · April 15th 2008, 10:41 AM

If you can't see how to factorize, you can use this:

$ax^2 + bx + c = (x - x_1)(x-x_2)$

$20x^2 - 11x - 3$

$x_1 = \frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{11+\sqrt{361}}{40} = \frac{3}{4}$

$x_2 = \frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{11-\sqrt{361}}{40} = -\frac{1}{5}$

$ax^2 + bx + c = \left(x - \frac{3}{4}\right)\left(x+\frac{1}{5}\right)$

Multiplying by 4.5 gives $4.\left(x - \frac{3}{4}\right).5.\left(x+\frac{1}{5}\right)$

$(4x-3)(5x+1)$

**red_dog** · April 15th 2008, 10:44 AM

Originally Posted by wingless

If you can't see how to factorize, you can use this:

$ax^2 + bx + c = (x - x_1)(x-x_2)$

$ax^2+bx+c=a(x-x_1)(x-x_2)$

**topsquark** · April 15th 2008, 11:12 AM

Originally Posted by angel.white

That's pretty cool, I actually didn't know about the ordered pair method.

Honestly I didn't learn it myself until I taught a Pre-Calc College class. It's called the "a-c" method. (The "a-c" comes from the coefficients in $ax^2 + bx + c$ .)

-Dan

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