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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!
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
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
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.
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.
Let’s say
f(y) = 3y2 - cos y + 1/y
Then f(x + 5)
= 3 * (x + 5)^2 - cos (x + 5) + 1/(x + 5)
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