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secondary 4 | A Maths
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Need help with 8a as I don't understand the question. Thx in advance!
Date Posted: 4 years ago
Views: 308
If the cubic polynomial has a factor of x, that means all terms in the polynomial are divisible by x. There aren't any constant terms.
Let the polynomial f(x) = ax³ + bx² + cx
= x(ax² + bx + c)
Now
f(x) - f(x - 1) = 3x² + 5x - 8
So ax³ + bx² + cx - ( a(x - 1)³ + b(x - 1)² + c(x - 1) ) = 3x² + 5x - 8
ax³ + bx² + cx - ( a(x³ - 3x² + 3x - 1) + b(x² - 2x + 1) + cx - c) = 3x² + 5x - 8
ax³ + bx² + cx - ax³ + 3ax² - 3ax + a - bx² + 2bx - b - cx + c = 3x² + 5x - 8
3ax² + (2b - 3a) x + a + c - b = 3x² + 5x - 8
Comparing coefficients,
3a = 3 → a = 1
2b - 3a = 5
Sub a = 1, 2b - 3(1) = 5
2b = 5 + 3 = 8
b = 4
a + c - b = -8
Sub a = 1, b = 4,
1 + c - 4 = -8
c = -8 + 4 - 1 = -5
So f(x) = x³ + 4x² - 5x
Remainder when f(x) is divided by x + 1
= f(-1)
= (-1)³ + 4(-1)² - 5(-1)
= -1 + 4 + 5
= 8
Let the polynomial f(x) = ax³ + bx² + cx
= x(ax² + bx + c)
Now
f(x) - f(x - 1) = 3x² + 5x - 8
So ax³ + bx² + cx - ( a(x - 1)³ + b(x - 1)² + c(x - 1) ) = 3x² + 5x - 8
ax³ + bx² + cx - ( a(x³ - 3x² + 3x - 1) + b(x² - 2x + 1) + cx - c) = 3x² + 5x - 8
ax³ + bx² + cx - ax³ + 3ax² - 3ax + a - bx² + 2bx - b - cx + c = 3x² + 5x - 8
3ax² + (2b - 3a) x + a + c - b = 3x² + 5x - 8
Comparing coefficients,
3a = 3 → a = 1
2b - 3a = 5
Sub a = 1, 2b - 3(1) = 5
2b = 5 + 3 = 8
b = 4
a + c - b = -8
Sub a = 1, b = 4,
1 + c - 4 = -8
c = -8 + 4 - 1 = -5
So f(x) = x³ + 4x² - 5x
Remainder when f(x) is divided by x + 1
= f(-1)
= (-1)³ + 4(-1)² - 5(-1)
= -1 + 4 + 5
= 8
I got the same expression for f(x), but I highly doubt that they are required to find the actual expression for f(x).
Thanks for the help, J and Eric! I don't think it was necessary to find equation f(x) in part a as it was only 2 marks, but it definitely did help with the other parts!
Welcome Annie.
Anyway, since it's only 2 marks for part a), this working is too long for that. But as mentioned in the other comment, it could come in useful for other questions. Just showing you how another method works for enrichment and broadening your learning :)
Anyway, since it's only 2 marks for part a), this working is too long for that. But as mentioned in the other comment, it could come in useful for other questions. Just showing you how another method works for enrichment and broadening your learning :)
See 2 Answers
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Good evening Annie! Here are my workings for this question.
This is an unusual question.
The fact that f(x) has a factor x (equivalently, a factor “x - 0”) means that when f(x) is divided by “x - 0”, the remainder is f(0) = 0 times other terms which remains zero.
Let me know if you need more explanation on this.
This is an unusual question.
The fact that f(x) has a factor x (equivalently, a factor “x - 0”) means that when f(x) is divided by “x - 0”, the remainder is f(0) = 0 times other terms which remains zero.
Let me know if you need more explanation on this.
Date Posted:
4 years ago
Note that the second equation f(x) - f(x - 1) = 3x2 + 5x - 8 holds true for any value of x, so we can choose x = 0 so that
f(0) - f(-1) = -8
where f(0) = 0 as I explained earlier.
So, 0 - f(-1) = -8, leaving f(-1) = 8.
This statement suggests that the remainder when f(x) is divided by x + 1 is -8.
f(0) - f(-1) = -8
where f(0) = 0 as I explained earlier.
So, 0 - f(-1) = -8, leaving f(-1) = 8.
This statement suggests that the remainder when f(x) is divided by x + 1 is -8.
What does f(x-1) mean? Does it mean substituting all the x in f(x) with "x-1"?
Yup!
Let’s say
f(y) = 3y2 - cos y + 1/y
Then f(x + 5)
= 3 * (x + 5)^2 - cos (x + 5) + 1/(x + 5)
Let’s say
f(y) = 3y2 - cos y + 1/y
Then f(x + 5)
= 3 * (x + 5)^2 - cos (x + 5) + 1/(x + 5)
But in this situation it’s difficult to find the actual function f(x) because we are only given that the difference between f(x) and f(x - 1) is 3x2 + 5x - 8. The original f(x) contains a factor x.
The remainder should be 8 instead of -8 in the working.
Anyway Annie you can take a look at an alternative approach I wrote in the main comments. Longer but might be useful in other questions
Anyway Annie you can take a look at an alternative approach I wrote in the main comments. Longer but might be useful in other questions
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Good evening Annie! Finding the actual expression f(x) is much more tedious because there are a lot of things to consider.
Date Posted:
4 years ago
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