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how did u get that?
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
im not sure
-15*4=-60 and -15+4=-11
So replace -11x=-15x+4x
so we have 10x^2-15x+4x-6
Factor by grouping
What do the first two terms have in common
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
You can also do trial factors like anfar if you like
i also have the problem 125n^2 + 50n + 5
ok so 5x(2x-3)+2(2x+3)
(2x+3)(5x+2) which is what anfar got
it was suppose to be 5x(2x-3)+2(2x-3)
now thats what anfar got
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.
would it be 5(5n+ 2)....?
no. it should be 5(5n+1)(5n+1)=5(5n+1)^2
but then what happens to the 10?
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
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?
thank you so much!
good.. u r welcome.