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secondary 3 | A Maths
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Hi, would really appreciate help to advise the working solution for this question. Thanks a lot :)
Date Posted: 3 years ago
Views: 325
Good afternoon Candice! The trick in doing such questions is to realise that a polynomial will leave a remainder of degree up to 1 when divided by a polynomial of degree 2, such that
F(x) = (2x^2 + 3x - 2) Q(x) + Ax + B.
F(x) = (2x - 1) (x + 2) Q(x) + Ax + B
where Ax + B is our remainder.
Since the polynomial is divisible by 2x - 1 but leaves a remainder of -5 when divided by x + 2,
f(1/2) = 0
0 = 1/2 A + B
B = -1/2 A
f(-2) = -5
-5 = -2A + B
-5 = -2A - 1/2 A
-5 = -5/2 A
A = 2
B = -1
So our remainder is 2x - 1.
F(x) = (2x^2 + 3x - 2) Q(x) + Ax + B.
F(x) = (2x - 1) (x + 2) Q(x) + Ax + B
where Ax + B is our remainder.
Since the polynomial is divisible by 2x - 1 but leaves a remainder of -5 when divided by x + 2,
f(1/2) = 0
0 = 1/2 A + B
B = -1/2 A
f(-2) = -5
-5 = -2A + B
-5 = -2A - 1/2 A
-5 = -5/2 A
A = 2
B = -1
So our remainder is 2x - 1.
This is our only approach to this question because we are unsure of what the polynomial at the start, f(x) could have been. In other words, we are unable to compare coefficients and so on.
Good afternoon Mr Eric. Thank you so much for taking time to provide me with such valuable advice. I really appreciate your patience and effort in going extra mile. I hope you have a good day ahead :D
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Thank you so much for your prompt advice and great effort in providing such neat & clear working solution, Mr Chau :) Really appreciated.
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