Question

Can anyone help me with factoring

Answer 1

6b^2-13b+3=-3

Answer 2

well here is a method that works on any quadratic that can be factored

start by rewriting the problem

\[6b^2 - 13b + 6 = 0\]

for any quadratic \[ax^2 + bx + c = 0\]

multiply a and c

so in your question multiply 6 by 6

then find the factors that add to -13.... both factors will be negative...

can you find the factors...

Answer 3

-9 and -4

Answer 4

@campbell_st

Answer 5

great so you now write the problem as

\[\frac{(ax + factor~1)(ax + factor~2)}{a}\]

so it becomes

\[\frac{(6x - 4)(6x - 9)}{6}\]

can you remove the common factors in each binomial... they will cancel with the denominator.

Answer 6

what is left will be the factored form of the quadratic... and oops just saw I changed b to x... sorry about that

Answer 7

it should be

\[\frac{(6b - 4)(6b - 9)}{6} = 0\]

Answer 8

ok

Answer 9

now i divide or multiply by 6

Answer 10

so to make it simple the binomials can be factored to

\[\frac{2(3x -2) 3(2x - 9)}{6} = 0\]

Answer 11

now just cancel the common factors... remember the numerator can be written as

\[\frac{2 \times 3\times(3x -2)(2x -3)}{6}\]

Answer 12

yea

Answer 13

so you have 2 linear factors that you can solve

(3x -2) = 0 and (2x -3) = 0

Answer 14

then add 2 and add 3

Answer 15

2/3 and 3/2

Answer 16

if you do that you get

3x = 2 and 2x = 3

find the values of x that make the equation true

Answer 17

now do i put them in paratheses

Answer 18

2/3 and 3/2

Answer 19

?