(20b(squared)-11b-3)

Sorry I don't know how to type the squared symbol...but thanks in advance!

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- Apr 15th 2008, 10:19 AMcurlysueFactoring and impossible equation!!!
(20b(squared)-11b-3)

Sorry I don't know how to type the squared symbol...but thanks in advance! - Apr 15th 2008, 10:26 AMangel.white
- Apr 15th 2008, 10:26 AMtopsquark
$\displaystyle 20b^2 - 11b - 3$

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.

$\displaystyle 20b^2 - 11b - 3 = 20b^2 + 4b - 15b - 3$

Group the terms as follows:

$\displaystyle = (20b^2 + 4b) + (-15b - 3)$

Now factor from each group:

$\displaystyle = 4b(5b + 1) + (-3)(5b + 1)$

and note that both terms have a common factor, 5b + 1, so we can factor that from each term as well:

$\displaystyle = (4b - 3)(5b + 1)$

There you go!

-Dan - Apr 15th 2008, 10:30 AMangel.white
- Apr 15th 2008, 10:31 AMred_dog
$\displaystyle 20b^2-11b-3=20b^2+4b-15b-3=4b(5b+1)-3(5b+1)=(5b+1)(4b-3)$

- Apr 15th 2008, 10:41 AMwingless
If you can't see how to factorize, you can use this:

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

$\displaystyle 20x^2 - 11x - 3$

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

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

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

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

$\displaystyle (4x-3)(5x+1)$ - Apr 15th 2008, 10:44 AMred_dog
- Apr 15th 2008, 11:12 AMtopsquark