f(x) = x³ - 2x² + ax + b, where a and b are constants - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 2
Question 7
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 7 - 2005 - Paper 2
Step 1
(a) Find the value of a and the value of b.
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Answer
To find the values of a and b, we can use the Remainder Theorem.
Given that when f(x) is divided by (x - 2), the remainder is 1, we can set up the equation:
f(2)=23−2(22)+2a+b=1
Simplifying, we get:
8−8+2a+b=12a+b=1ag1
From the second condition, when f(x) is divided by (x + 1), the remainder is 28:
f(−1)=(−1)3−2(−1)2+a(−1)+b=28
Simplifying, we get:
−1−2−a+b=28−a+b=31ag2
Now we solve the two equations (1) and (2):
From (1):
b=1−2a
Substituting this into (2):
−a+(1−2a)=31
Simplifying gives:
1−3a=31−3a=30a=−10
Plugging the value of a back into equation (1):
2(−10)+b=1−20+b=1b=21
Thus, the values are:
a=−10
b=21
Step 2
(b) Show that (x - 3) is a factor of f(x).
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Answer
To show that (x - 3) is a factor of f(x), we need to evaluate f(3) and check if the result is zero:
Substitute x = 3 into f(x):
f(3)=33−2(32)+a(3)+b
Substituting the known values of a and b:
=27−18+(−10)(3)+21=27−18−30+21=0
Since f(3) = 0, it follows that (x - 3) is indeed a factor of f(x).
