Question

10x^2 - 11x - 6 Factor

Answer 1

(5x+2)(2x-3)

Answer 2

how did u get that?

Answer 3

This is in the form of ax^2+bx+c where a=10, b=-11, c=-6. Find factors of a*c that add up to be b. a*c=(10)(-6)=-60. What are factors of -60 that add up to be -11

Answer 4

im not sure

Answer 5

-15*4=-60 and -15+4=-11

Answer 6

So replace -11x=-15x+4x

Answer 7

so we have 10x^2-15x+4x-6

Answer 8

Factor by grouping

Answer 9

What do the first two terms have in common

Answer 10

2?

Answer 11

this literally by observation.
10= 1*10= 2*5
factor of 6: 1*6=2*3
try mix and match the factors in form
(ax+b)(cx+d)
where a&c is a pair of factor 10 and b&d a pair of factor 6.
and ad+bc=-11

Answer 12

o okay

Answer 13

You can also do trial factors like anfar if you like

Answer 14

i also have the problem 125n^2 + 50n + 5

Answer 15

ok so 5x(2x-3)+2(2x+3)

Answer 16

(2x+3)(5x+2) which is what anfar got

Answer 17

oops

Answer 18

it was suppose to be 5x(2x-3)+2(2x-3)

Answer 19

=(5x+2)(2x-3)

Answer 20

now thats what anfar got

Answer 21

for the 2nd problem:

first, make the numbers smaller.note that 125,50 and 5 have common factors 5.
so now u have:
5(25n^2+10n+1)
then, u can try use the mix and match for the equation in the bracket.

Answer 22

would it be 5(5n+ 2)....?

Answer 23

no. it should be 5(5n+1)(5n+1)=5(5n+1)^2

Answer 24

but then what happens to the 10?

Answer 25

i put it this way:
let say u have an equation in form Ax^2+Bx+C:
u have to factor it in form of (ax+b)(cx+d)
where:
a*c=A , b*d=C and ad+bc=B

Answer 26

okay

Answer 27

in the 2nd case, u considering 25n^2+10n+1
so, u hav A=25, B=10 and C=1
note that 1=1*1 or -1*-1

since B is positive, u take the positive ones. so, u now have b=d=1
thus,
a*c=25 and a+c=10

now, 25=25*1=5*5.
25+1 not equal to 10. so u take a=c=5.
therefore u have:
5(5n+1)(5n+1)=5(5n+1)^2 as the answer.
u got it?

Answer 28

yess

Answer 29

thank you so much!

Answer 30

good.. u r welcome.