# IBDP Maths analysis and approaches Topic: AHL 2.12 :The factor and remainder theorems HL Paper 1

### Question

The polynomial $$P(x) = {x^3} + a{x^2} + bx + 2$$ is divisible by (x +1) and by (x − 2) .

Find the value of a and of b, where $$a,{\text{ }}b \in \mathbb{R}$$ .

### Markscheme

METHOD 1

As (x +1) is a factor of P(x), then P(−1) = 0     (M1)

$$\Rightarrow a – b + 1 = 0$$ (or equivalent)     A1

As (x − 2) is a factor of P(x), then P(2) = 0     (M1)

$$\Rightarrow 4a + 2b + 10 = 0$$ (or equivalent)     A1

Attempting to solve for a and b     M1

a = −2 and b = −1     A1     N1

[6 marks]

METHOD 2

By inspection third factor must be x −1.     (M1)A1

$$(x + 1)(x – 2)(x – 1) = {x^3} – 2{x^2} – x + 2$$     (M1)A1

Equating coefficients a = −2, b = −1     (M1)A1     N1

[6 marks]

METHOD 3

Considering $$\frac{{P(x)}}{{{x^2} – x – 2}}$$ or equivalent     (M1)

$$\frac{{P(x)}}{{{x^2} – x – 2}} = (x + a + 1) + \frac{{(a + b + 3)x + 2(a + 2)}}{{{x^2} – x – 2}}$$     A1A1

Recognising that $$(a + b + 3)x + 2(a + 2) = 0$$     (M1)

Attempting to solve for a and b     M1

a = −2 and b = −1     A1     N1

[6 marks]

### Question

When $$f(x) = {x^4} + 3{x^3} + p{x^2} – 2x + q$$ is divided by (x − 2) the remainder is 15, and (x + 3) is a factor of f(x) .

Find the values of p and q .

### Markscheme

$$f(2) = 16 + 24 + 4p – 4 + q = 15$$     M1

$$\Rightarrow 4p + q = – 21$$     A1

$$f( – 3) = 81 – 81 + 9p + 6 + q = 0$$     M1

$$\Rightarrow 9p + q = – 6$$     A1

$$\Rightarrow p = 3{\text{ and }}q = – 33$$     A1A1     N0

[6 marks]

### Question

When the function $$q(x) = {x^3} + k{x^2} – 7x + 3$$ is divided by (x + 1) the remainder is seven times the remainder that is found when the function is divided by (x + 2) .

Find the value of k .

### Markscheme

$$q( – 1) = k + 9$$     M1A1

$$q( – 2) = 4k + 9$$     A1

$$k + 9 = 7(4k + 9)$$     M1

$$k = – 2$$     A1

Notes: The first M1 is for one substitution and the consequent equations.

Accept expressions for $$q( – 1)$$ and $$q( – 2)$$ that are not simplified.

[5 marks]

### Question

When $$3{x^5} – ax + b$$ is divided by x −1 and x +1 the remainders are equal. Given that a , $$b \in \mathbb{R}$$ , find

(a)     the value of a ;

(b)     the set of values of b .

### Markscheme

(a)     $$f(1) = 3 – a + b$$     (A1)

$$f( – 1) = – 3 + a + b$$     (A1)

$$3 – a + b = – 3 + a + b$$     M1

$$2a = 6$$

$$a = 3$$     A1     N4

(b)     b is any real number     A1

[5 marks]

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