Question

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Answer 1

@Jhannybean

Answer 2

Heyo. whats the problem?

Answer 3

can you help me factor a trinomial

Answer 4

Yeah sure, whats the trinomial?

Answer 5

$$ \Large 8a^2 -18a -5 = 0 $$

Answer 6

\[8a^2 + 2a - 20a - 5\]
factorise it.....

Answer 7

Yeah theres that way too haha

Answer 8

yep that works. factoring by grouping

Answer 9

Then theres this way for trinomials you cannot factor by grouping.

Answer 10

ya

Answer 11

Let me know where my steps dont make sense, :) \[\color{red}{8}a^2-18a\color{red}{-5}=0\]\[a^2-18a\color{red}{-40}=0\]\[(a-20)(a+2)=0\]NOw that you've simplified the polynomial into a simple quadratic so you can factor it, use the leading coefficient to redivide the factor so you can stay consistent with the original polynomial \[\left(a-\frac{20}{8}\right)\left(a+\frac{2}{8}\right)=0\]\[\left(a-\frac{5}{2}\right)\left(a+\frac{1}{4}\right)=0\]Now that your factors are completely simplified, you multiply the denominator of each factor, i.e the 2 and 4, to the variables to further simplify the fractional form. The way I show it is a shortcut, but the proper way to show it would be \[\left(\frac{2a-5}{2}\right)\left(\frac{4a+1}{4}\right)=0\]\[\left(\frac{2a-5}{\color{red}{2}}\right)\left(\frac{4a+1}{\color{red}{4}}\right)=\color{red}{0}\]The factors would now multiply eachother, and the value in the denominator gets multiplied to the 0. This becomes \[(2a-5)(4a+1)=0\]

Answer 12

Generally works for trinomials you cannot factor by grouping, I guess.

Answer 13

actually i think that if your method works, so does grouping.

Answer 14

are there any requarements what method to use?

Answer 15

i mean , if you can't factor by grouping then you won't be able to factor that way

Answer 16

No not really, I just test and check between grouping, completing the square and this method, called the box method.

Answer 17

well one of the easiest methods is quadratic formula.
8a^2-18a-5=0

use quadratic formula: (18+-sqrt(324+160))/16=(18+-22)/16=2.5 and -0.25.

then factored form is 8*(x-2.5)*(x+0.25)

grouping method is also one of methods, just longer to do in this case.

Answer 18

i'm refering to this comment
"Generally works for trinomials you cannot factor by grouping, I guess."

Answer 19

Oh, hmm... I've tried factoring some trinomials before that I couldnt expand by grouping... and this method helped. Maybe I'm mistaken?

Answer 20

I mean, your method does not work for trinomials you cannot factor by grouping.
for example
$$ \Large { 2x^2 + 2x -5 } $$

We can't factor by grouping or use your method

Answer 21

Let's see

Answer 22

well there are soooo many different method to factor it
and if leading coefficient isn't one then i like this method to factor quadratic equation ;P
factor first and last term
cross multiply and if you get middle term then
(4a+1)( 2a -5) factor of 8a^2-18-5