Solve this question class 9 polynomials
Attachments:
Answers
Answered by
3
If polynomial f(x) is divisible by (x-1) and (x+1), it means that (x-1) and (x+1) are one of the factors of the polynomial.
Now, let's solve for the (x-1) factor [x - 1 = 0 => x = 1],
f(x) = x^4 - 2x^3 + 3x^2 -ax + b
f(1) = (1)^4 - 2(1)^3 + 3(1)^2 - a(1) + b
= 1 - 2 + 3 - a + b
= 2 + b - a
Also, given remainder = 5
therefore, 2+b-a = 5
b - a = 3 ......... (1)
Now, for the other factor (x+1) [x+1 = 0 => x=-1],
f(-1) = (-1)^4 - 2(-1)^3 + 3(-1)^2 - a(-1) + b
= 1 + 2 +3 + a + b
Also, given remainder = 19
therefore, 6 + a + b = 19
and, a + b = 13 ....... (2)
Now, to solve (1) and (2),
add equation (1) and (2),
b - a + a + b = 13 + 3
=> 2b = 16
=> b = 8
and put the value of the b obtained above in equation (2),
=> a + b = 13
=> a + 8 = 13
=> a = 5
therefore, b = 8 & a = 5
Now, let's solve for the (x-1) factor [x - 1 = 0 => x = 1],
f(x) = x^4 - 2x^3 + 3x^2 -ax + b
f(1) = (1)^4 - 2(1)^3 + 3(1)^2 - a(1) + b
= 1 - 2 + 3 - a + b
= 2 + b - a
Also, given remainder = 5
therefore, 2+b-a = 5
b - a = 3 ......... (1)
Now, for the other factor (x+1) [x+1 = 0 => x=-1],
f(-1) = (-1)^4 - 2(-1)^3 + 3(-1)^2 - a(-1) + b
= 1 + 2 +3 + a + b
Also, given remainder = 19
therefore, 6 + a + b = 19
and, a + b = 13 ....... (2)
Now, to solve (1) and (2),
add equation (1) and (2),
b - a + a + b = 13 + 3
=> 2b = 16
=> b = 8
and put the value of the b obtained above in equation (2),
=> a + b = 13
=> a + 8 = 13
=> a = 5
therefore, b = 8 & a = 5
Saivi1:
plz solve it fully
its not helpful to me
solve it full nd i mark it brainliest
Okay, I tried to elaborate it more.
thnx
Mark Brainliest if it helped you, it means alot to us :) Have a good time!
i am not seeing any brainliest option
where it is gone i dont know
but tomorrow is my exam and thanks for your help ..
It must be there, try to look for it! And good luck with your exams, hope you ace it!
Similar questions