At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
Q(x) is an unknown polynomial?
then... a = -1, b = 2?...dont think it should be this simple
that isnt the answer lol
answer is a = 20 b = 24
but i cant get it :(
if Q(x) = 5x-9 ?
o..o maybe you could work out the full solution and see if it matches my answer :P
\[ (x+1)*(x+2)*(5*x-9) = 5x^3 + 6x^2 -17x - 18 + 16x +20=5x^3 + 6x^2 -x + 2\]
oo then how should i carry on
hmm a is 16 and b is 20 for me
check for any careless mistake? :O
no i don't think so
can you upload the picture of your working?
i didn't do it on my notebook, in head.
\[(x+1)(x+2)(5x−9) +ax + b = 5x^3+6x^2−17x−18+ ax+b\]
i did it also and got same result as ishaan
i need the working :(
or rather simple explanation towards the steps
equating coefficients of x gives -1 = 10 - 27 + a a = 27-11 = 16
use q(x) as px +q, suggested by you earlier if you really need to show working
can i let Q(x) = px + q will i be able to get the answer?
oh but i cant seem to get the answer somehow :(
i get decimal numbers :(
show me your working i will try to guide you
kk i try again
well you can get multiple answers, maybe.
only a and b
(x+1) (x-2) *********
Ishaan i edited the question
sry about it
i can see how ishaan got p = 5 because of the 5x^3 but the q = -9 is not obvious = 5x^3 + 15x^2 + 10x + qx^2 + 3qx + 2q + ax + b = 5x^3 + (15 + q)x^2 + (10 + 3q + a)x + 2q + b compare coefficients with 5x^3 + 6x^2 - x + 2 of x^2: q + 15 = 6 so q = -9 of x: 10 + a + 3q = 10 + a - 27 = a - 17 and a -17 = -1 so a = 16 of x^0 : 2q + b = 2 so -18 + b = 2 so b = 20 a=16, b = 20