6a^2+5a-4
factor..someone plz?

Question

6a^2+5a-4
factor..someone plz?

jim_thompson5910 · Answer

6a^2+5a-4


6a^2+8a-3a-4


(6a^2+8a)+(-3a-4)


Keep going to completely factor

anonymous · Answer

is (12a+20a) (-3a-7) the final answer??

jim_thompson5910 · Answer

no

anonymous · Answer

Then wat is it?? :S

jim_thompson5910 · Answer

what can you factor out of 6a^2+8a

anonymous · Answer

a 2??

jim_thompson5910 · Answer

close, more like 2a

jim_thompson5910 · Answer

6a^2+8a = 2a(3a + 4)

anonymous · Answer

(3a-4)(a+1)
a=4/3 or a=-1

jim_thompson5910 · Answer

now, is it possible to write (-3a-4) in the form k(3a+4) ?

anonymous · Answer

i dont know.... you tell me?There are no k's so y wud u add one??? :s

jim_thompson5910 · Answer

k can be any number or expression, think of it as a variable

anonymous · Answer

ohh ok

jim_thompson5910 · Answer

ok let me ask this

is it possible to factor (-3a-4) to get (3a+4) times SOMETHING where that SOMETHING is an expression or number ??

anonymous · Answer

huh???

jim_thompson5910 · Answer

is it possible to turn (-3a-4) into (3a+4) ?

campbell_st · Answer

here is a really simple method for any difficult factorisation 

if the quadratic is                                                   Your question

ax^2 + bx + c = 0                                             6a^2 + 5a - 4 = 0

1. multiply a and c                                              6 x -4 = -24

2. Find the factors of a*c that add to b                 factors   -3 + 8 = 5  

next step up a fraction with a as the denominator 

the numerator is in the form  (ax + factor 1)(ax + factor 2)

for your question 

|dw:1351990505786:dw|

this meothod works everytime... you don't need to split the middle term...