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Mathematics 23 Online
OpenStudy (anonymous):

Given that 5x^3 + 6x^2 -x + 2 = (x+1) (x-2) Q(x) + ax + b find the value of a and b. Question on Identities :)

OpenStudy (anonymous):

Q(x) is an unknown polynomial?

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

then... a = -1, b = 2?...dont think it should be this simple

OpenStudy (anonymous):

that isnt the answer lol

OpenStudy (anonymous):

answer is a = 20 b = 24

OpenStudy (anonymous):

but i cant get it :(

OpenStudy (anonymous):

if Q(x) = 5x-9 ?

OpenStudy (anonymous):

o..o maybe you could work out the full solution and see if it matches my answer :P

OpenStudy (anonymous):

Thank you.

OpenStudy (anonymous):

\[ (x+​1)*​(x+​2)*​(5*​x-​9) = 5x^3 + 6x^2 -17x - 18 + 16x +20=5x^3 + 6x^2 -x + 2\]

OpenStudy (anonymous):

@Sgstudent

OpenStudy (anonymous):

oo then how should i carry on

OpenStudy (anonymous):

hmm a is 16 and b is 20 for me

OpenStudy (anonymous):

check for any careless mistake? :O

OpenStudy (anonymous):

no i don't think so

OpenStudy (anonymous):

can you upload the picture of your working?

OpenStudy (anonymous):

i didn't do it on my notebook, in head.

OpenStudy (anonymous):

\[(x+​1)(x+​2)(5x−​9) +ax + b = 5x^3+6x^2−17x−18+ ax+b\]

OpenStudy (cwrw238):

i did it also and got same result as ishaan

OpenStudy (anonymous):

i need the working :(

OpenStudy (anonymous):

or rather simple explanation towards the steps

OpenStudy (cwrw238):

equating coefficients of x gives -1 = 10 - 27 + a a = 27-11 = 16

OpenStudy (anonymous):

use q(x) as px +q, suggested by you earlier if you really need to show working

OpenStudy (anonymous):

can i let Q(x) = px + q will i be able to get the answer?

OpenStudy (anonymous):

oh but i cant seem to get the answer somehow :(

OpenStudy (anonymous):

i get decimal numbers :(

OpenStudy (anonymous):

show me your working i will try to guide you

OpenStudy (anonymous):

kk i try again

OpenStudy (anonymous):

well you can get multiple answers, maybe.

OpenStudy (anonymous):

only a and b

OpenStudy (anonymous):

(x+1) (x-2) *********

OpenStudy (anonymous):

Ishaan i edited the question

OpenStudy (anonymous):

sry about it

OpenStudy (cwrw238):

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

OpenStudy (anonymous):

Thanks :)

OpenStudy (cwrw238):

yw

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