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secondary 4 | A Maths
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urgent help pls
Since the coefficient of the term with the highest power of x (x³) is not known, let it be A.
Let f(x) = A(x² - x - 2)(x - k)
= A(x - 2)(x + 1)(x - k)
Since the remainders are 20 and -100 when f(x) is divided by (x - 1) and (x + 2) respectively,
By the remainder theorem, f(1) = 20
A(1 - 2)(1 + 1)(1 - k) = 20
-2A(1 - k) = 20
2A(k - 1) = 20
A = 10/(k - 1) ①
By the remainder theorem, f(-2) = -100
A(-2 - 2)(-2 + 1)(-2 - k) = -100
4A(-2 - k) = -100
4A(2 + k) = 100
A = 25/(2 + k) ②
① = ②
10/(k - 1) = 25/(2 + k)
Cross multiply,
10(2 + k) = 25(k - 1)
20 + 10k = 25k - 25
15k = 45
k = 3
Then, A = 10/(3 - 1) = 10/2 = 5
Or, A = 25/(2 + 3) = 25/5 = 5
So,
f(x) = 5(x² - x - 2)(x - 3)
= 5(x³ - 3x² - x² + 3x - 2x + 6)
= 5(x³ - 4x² + x + 6)
= 5x³ - 20x² + 5x + 30
g(x) = f(x) - (x - 3)
= 5(x² - x - 2)(x - 3) - (x - 3)
= (x - 3)(5(x² - x - 2) - 1)
= (x - 3)(5x² - 5x - 10 - 1)
= (x - 3)(5x² - 5x - 11)
Since g(x) is also a multiple of (x - 3), (or in other words, (x - 3) is a factor of g(x))
g(x) is divisible by (x - 3) , so there is only a quotient of 5x² - 5x - 11 and no remainder.
We deduce that the Remainder = 0
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