A-Level Edexcel Maths Pure Circles: f(x) = x^3 + ax^2 + bx + 3, wher

f(x) = x^3 + ax^2 + bx + 3, where a and b are constants - Edexcel - A-Level Maths Pure - Question 7 - 2012 - Paper 4

Question 7

f(x) = x^3 + ax^2 + bx + 3, where a and b are constants. Given that when f(x) is divided by (x+2) the remainder is 7, (a) show that 2a - b = 6. Given also that wh... show full transcript

Worked Solution & Example Answer:f(x) = x^3 + ax^2 + bx + 3, where a and b are constants - Edexcel - A-Level Maths Pure - Question 7 - 2012 - Paper 4

Step 1

(a) show that 2a - b = 6

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Answer

To find the value of the remainder when dividing by (x + 2), we use the Remainder Theorem, which states that the remainder of f(x) when divided by (x - r) is f(r).

Calculating f(-2):

$f(-2) = (-2)^3 + a(-2)^2 + b(-2) + 3$ $= -8 + 4a - 2b + 3$ $= -5 + 4a - 2b$

Setting this equal to the remainder 7:

$-5 + 4a - 2b = 7$

Solving for a and b gives:

$4a - 2b = 12$

Dividing the entire equation by 2:

$2a - b = 6$

Step 2

(b) find the value of a and the value of b.

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Answer

Using the information that the remainder when f(x) is divided by (x - 1) is 4, we apply the Remainder Theorem again:

Calculating f(1):

$f(1) = 1^3 + a(1)^2 + b(1) + 3$ $= 1 + a + b + 3$ $= 4 + a + b$

Setting this equal to the remainder 4:

$4 + a + b = 4$

This simplifies to:

$a + b = 0$

Now we have the system of equations:

$2a - b = 6$
$a + b = 0$

Substituting b from the second equation into the first: $2a - (-a) = 6$ $2a + a = 6$ $3a = 6$ $a = 2$

Substituting back to find b: $b = -a = -2$

Thus, the values are: $a = 2$ and $b = -2$ .

f(x) = x^3 + ax^2 + bx + 3, where a and b are constants - Edexcel - A-Level Maths Pure - Question 7 - 2012 - Paper 4

Worked Solution & Example Answer:f(x) = x^3 + ax^2 + bx + 3, where a and b are constants - Edexcel - A-Level Maths Pure - Question 7 - 2012 - Paper 4

(a) show that 2a - b = 6

(b) find the value of a and the value of b.

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