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secondary 3 | A Maths
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Hi, can someone help me out with qn 3, please?
Date Posted: 2 years ago
Views: 435
alternative solution for (ii), without finding polynomial Q(x)
sub x = 1,
1+a+b-3 = 0 x Q(1) - 1 - 2
gives a+b = -1 ...... (1)
sub x = -1
-1-a+b-3 = 0 x Q(-1) + 1 - 2
gives -a+b = 3 ...... (2)
solve (1) & (2) simultaneously,
gives a = -2 , b = 1
(matches Nicholas' answer below)
sub x = 1,
1+a+b-3 = 0 x Q(1) - 1 - 2
gives a+b = -1 ...... (1)
sub x = -1
-1-a+b-3 = 0 x Q(-1) + 1 - 2
gives -a+b = 3 ...... (2)
solve (1) & (2) simultaneously,
gives a = -2 , b = 1
(matches Nicholas' answer below)
See 2 Answers
done
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part i and ii
Date Posted:
2 years ago
for part i, why is the degree 3?
Thank you for helping me out!
Hello, good question! The degree of polynomial refers to the highest power of x in this case. Here, we see that Q(x) is multiplied by (x^2-1) to get x^5 on the left side. Hence, in order to get x^5, Q(x) would have a x^3 present and this would be the degree as it’s the highest power. [x^3 * x^2 is x^5)
Ahhhh....now i understand. Thank you so much, Tutor!
done
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part iii, you can use long division as well
Date Posted:
2 years ago
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