f(x) = x³ - 2x² + ax + b, where a and b are constants - Edexcel - A-Level Maths Pure - Question 6 - 2005 - Paper 2
Question 6
f(x) = x³ - 2x² + ax + b, where a and b are constants.
When f(x) is divided by (x - 2), the remainder is 1.
When f(x) is divided by (x + 1), the remainder is 28.
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Worked Solution & Example Answer:f(x) = x³ - 2x² + ax + b, where a and b are constants - Edexcel - A-Level Maths Pure - Question 6 - 2005 - Paper 2
Step 1
Find the value of a and the value of b.
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Answer
To find the constants a and b, we can use the Remainder Theorem, which states that the remainder of the polynomial f(x) when divided by (x - c) is given by f(c).
First, we know that:
When f(x) is divided by (x - 2), the remainder is 1. Therefore, we have:
f(2)=23−2(2)2+2a+b=1
This simplifies to:
8−8+2a+b=12a+b=1(1)
Next, when f(x) is divided by (x + 1), the remainder is 28. Thus:
f(−1)=(−1)3−2(−1)2+(−1)a+b=28
This simplifies to:
−1−2−a+b=28−a+b=31(2)
Now we have a system of equations:
From (1):
2a+b=1
From (2):
−a+b=31
We can eliminate b by subtracting (1) from (2):
(−a+b)−(2a+b)=31−1−3a=30⇒a=−10
Substitute a back into (1):
2(−10)+b=1⇒−20+b=1⇒b=21
Thus, we find: a = -10 and b = 21.
Step 2
Show that (x - 3) is a factor of f(x).
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Answer
To determine whether (x - 3) is a factor of f(x), we evaluate f(3):
First, substitute x = 3 into f(x):
f(3)=33−2(3)2+a(3)+b
Substituting in the values of a and b gives:
f(3)=27−18+3(−10)+21
Simplifying:
f(3)=27−18−30+21⇒f(3)=0
Since f(3) = 0, this confirms that (x - 3) is indeed a factor of f(x).
