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f(x) = ax^3 + bx^2 - 4x - 3, where a and b are constants - Edexcel - A-Level Maths Pure - Question 4 - 2013 - Paper 6
Question 4
f(x) = ax^3 + bx^2 - 4x - 3, where a and b are constants. Given that (x - 1) is a factor of f(x), (a) show that a + b = 7 Given also that, when f(x) is divi... show full transcript
Worked Solution & Example Answer:f(x) = ax^3 + bx^2 - 4x - 3, where a and b are constants - Edexcel - A-Level Maths Pure - Question 4 - 2013 - Paper 6
Step 1
(a) show that a + b = 7
Answer
To show that , we can use the Factor Theorem. Since (x - 1) is a factor of f(x), it must satisfy the equation f(1) = 0.
Calculating f(1):
f(1) & = a(1)^3 + b(1)^2 - 4(1) - 3 \\ & = a + b - 4 - 3 \\ & = a + b - 7. directly setting this equal to zero gives us: $$a + b - 7 = 0$$ Thus, we have: $$a + b = 7.$$ This concludes the proof.Step 2
(b) find the value of a and the value of b, showing each step in your working.
Answer
Knowing that f(-2) = 9 when dividing f(x) by (x + 2), we can calculate f(-2):
f(-2) & = a(-2)^3 + b(-2)^2 - 4(-2) - 3 \\ & = -8a + 4b + 8 - 3 \\ & = -8a + 4b + 5. directly setting this equal to 9: $$-8a + 4b + 5 = 9$$ Thus, simplifying gives: $$-8a + 4b = 4$$ which we can divide by 4: $$-2a + b = 1.$$ Now we have a system of equations: 1. $a + b = 7$ 2. $-2a + b = 1$ Next, we can substitute the first equation into the second: $$-2a + (7 - a) = 1$$ Simplifying gives: $$-2a + 7 - a = 1$$ $$-3a + 7 = 1$$ $$-3a = -6$$ $$a = 2.$$ Now substituting $a = 2$ back into the first equation: $$2 + b = 7$$ $$b = 5.$$ Thus, the values of a and b are: $$a = 2$$ $$b = 5$$.