Can you explain to me how to factorise 12a^2 + 11a - 15?

Question

Can you explain to me how I would facorise 12a^2 + 11a - 15.

Answer 1

multiply 12 and -15 that is -180
factor -180 into 2 numbers such that sum is 11, (20 , -9) by trial and error. 15 , -12 sum is 3 so try >15 18,-10 no 20 -9 yesy

so 12a^2+11a-15
= 12a^2+20a -9a - 15
= 4a(3a+5) -3(3a+5)
= (3a+5)(4a-3)

Answer 2

The best to factorise a quadratic equation is to split the middle term. To do that, first multiply the coefficient of the first term with the last term.
12*-15 = -180
The middle term must be split into two terms.
The product of the coefficients of the two terms must equal -180.
It is obvious that one of them is negative. The smaller one must be negative because the two added equals 11, which is positive.
The numbers -9 and 20 seem to work.
20-9 = 11
Now let's get on with the work:
12a^2 + 11a - 15 = 12a^2 - 9a + 20a - 15
= 3a(4a - 3) + 5(4a - 3)
= (3a + 5)(4a - 3)

Answer 3

do 12 x 15 =180 thn find 2 nos that x to give this bt ad 2 give -11
= -20 and 9 then write out the equation again with -20 and 9 instead of 11a

12a^2 -20a +9a -15 then factorise the first alf of the eqnt then the 2nd part

4a(3a -5) +3 (3a -5) (you knw uve dun it rit whn both brackets r the same)

then its simply (3a-5) (4a+3)

Answer 4

here's the easy way 12a^2 + 11a - 15 =0

solve the equation by formula

>>>>>> a= 3/4 or -5/3

(a-3/4)(a+5/3)

{4(a-3/4)}{3(a+5/3)}

(4a-3)(3a+5) =12a^2 + 11a - 15

hit the site below to learn more about how to factorize
quadratics the standard way
http://www.maths.tcd.ie/~mc/Downloads/QuadraticCubic.pdf#search='factorise%20quadratic%20equations'

analysis:

(pa+q)(ra+s).............(1)
=pra^2+(sp+qr)a+qs..(2)
sp+qr=11...................(3)
prsq= -180.................(4)
combine (3)and(4)
-180/qr+qr=11
>>>qr = -9 or20
pr=12 from (2)
q/p=5/3
ps= -9, from(4)
r/s= -4/3
substitute back into (1)
(pa+5/3p)(ra-3/4r)
=pr(a+5/3)(a-3/4)
={3(a+5/3)}{4(a-3/4)}
=(3a+5)(4a-3) as required
(pr=12 from (2)=3*4)

i hope that this helps

Answer 5

Just mull the factors of 15 and the factors of 12 around in your head for a while until 5*4 minus 3*3 looks like it could give 11 i.e. (3a+5)(4a-3). You could use the formula for roots of a quadratic but that is a bit tedious.

Answer 6

ax^2 + bx + c
can be factorised into
(mx+p)(nx+q)
then: mn = a
pq=c
and pn+mq=b

So... lets assume there are whole number solutions. in this case m and n are either 3 and 4 or 2 and 6; and p and q are either +3 and -5 or -3 and +5.

With a bit of trial and error, and a bit of instinct, I know that our 11x can be produced from 20x - 9x
So we get a factorised polynomial of:
(3x + 5)(4x - 3)

Answer 7

12a^2+20a -9a - 15 = 4a(3a+5) -3(3a+5)= (3a+5)(4a-3)
is it alright???