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secondary 4 | A Maths
2 Answers Below
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please help thankyou
Date Posted: 5 years ago
Views: 235
Alternative method to part a) :
When divisor is (x-1) , remainder is f(1) = 12. Let the quotient be A.
When divisor is (x-4) , remainder is f(4) = 3.
Let the quotient be B.
dividend = quotient x divisor + remainder
So f(x) = B(x - 4) + 3 ①
f(x) also = A(x - 1) + 12
= (A(x - 4) + 3A + 9) + 3 ②
So (3A + 9) is divisible by (x - 4) and is therefore a multiple of (x - 4) since we can end up with a remainder of 3.
Compare ① and ②,
B(x - 4) = A(x-4) + 3A + 9
3A + 9 = (B - A)(x - 4)
3A = (x - 4)(B-A) - 9
A = ⅓(x - 4)(B - A) - 3
Sub this into f(x) = A(x - 1) + 12 ,
f(x) = (⅓(x - 4)(B - A) - 3)(x - 1) + 12
f(x) = ⅓(B - A)(x - 4)(x -1) - 3(x - 1) + 12
= ⅓(B - A)(x - 4)(x -1) + 15 - 3x
As you can see, ⅓(B-A) is the quotient and (x - 4)(x -1) is the divisor.
The 15 - 3x cannot be factorised into in terms of (x - 4)(x-1) as it is not divisible by that.
So the remainder is 15 - 3x.
(Note that A and B are not necessarily integers. They can be polynomials too.)
When divisor is (x-1) , remainder is f(1) = 12. Let the quotient be A.
When divisor is (x-4) , remainder is f(4) = 3.
Let the quotient be B.
dividend = quotient x divisor + remainder
So f(x) = B(x - 4) + 3 ①
f(x) also = A(x - 1) + 12
= (A(x - 4) + 3A + 9) + 3 ②
So (3A + 9) is divisible by (x - 4) and is therefore a multiple of (x - 4) since we can end up with a remainder of 3.
Compare ① and ②,
B(x - 4) = A(x-4) + 3A + 9
3A + 9 = (B - A)(x - 4)
3A = (x - 4)(B-A) - 9
A = ⅓(x - 4)(B - A) - 3
Sub this into f(x) = A(x - 1) + 12 ,
f(x) = (⅓(x - 4)(B - A) - 3)(x - 1) + 12
f(x) = ⅓(B - A)(x - 4)(x -1) - 3(x - 1) + 12
= ⅓(B - A)(x - 4)(x -1) + 15 - 3x
As you can see, ⅓(B-A) is the quotient and (x - 4)(x -1) is the divisor.
The 15 - 3x cannot be factorised into in terms of (x - 4)(x-1) as it is not divisible by that.
So the remainder is 15 - 3x.
(Note that A and B are not necessarily integers. They can be polynomials too.)
thankyou so much!!
Welcome
See 2 Answers
done
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The first one is not easy unless you know what is going on
Date Posted:
5 years ago
The degree of the remainder is at most 1 lower than the degree of the divisor. So it isn't always going to be 1 lower.
Sometimes there is no coefficient of x. In this case assuming that the form ax + b works since if it was actually two degrees lower, the remainders would be the same for both cases. It may not work for other questions and it is best to not assume so
Sometimes there is no coefficient of x. In this case assuming that the form ax + b works since if it was actually two degrees lower, the remainders would be the same for both cases. It may not work for other questions and it is best to not assume so
Should be ok I think, because if indeed the remainder does not contain x then a = 0.
But in this case a definitely cannot be zero since the remainders when divided by both numbers are different.
But in this case a definitely cannot be zero since the remainders when divided by both numbers are different.
thankyou so much!
done
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Second one
Date Posted:
5 years ago
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