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A-Level Pure 2 · Edexcel IAL WMA12 & CAIE 9709 · Topic pack

Topic pack · Polynomials แพ็กฝึกเฉพาะหัวข้อ · พหุนาม

10 questions 10 ข้อ 55 marks 55 คะแนน 70 min 70 นาที Factor & remainder theorem, easy → hard

Ten questions on one topic, ordered easy → hard. The last three are full exam difficulty. Open the mark schemes only after a real attempt. สิบข้อหัวข้อเดียว เรียงง่าย → ยาก สามข้อสุดท้ายคือระดับข้อสอบจริง เปิดมาร์คสกีมหลังลองทำจริงเท่านั้น

⏱ 70:00 Scoreคะแนนรวม: –
Q1. 3 marks

Find the remainder when $2x^3-5x^2+x-8$ is divided by $(x-2)$.

Mark schemeshow ▾
M1 remainder theorem: evaluate at $x=2$  M1 $16-20+2-8$  A1 $=-10$.
Q2. 4 marks

Show that $(x+1)$ is a factor of $f(x)=x^3-7x-6$, and factorise $f(x)$ completely.

Mark schemeshow ▾
M1 $f(-1)=-1+7-6$  A1 $=0$, so $(x+1)$ is a factor  M1 $f(x)=(x+1)(x^2-x-6)$  A1 $=(x+1)(x-3)(x+2)$.
Q3. 4 marks

Find the remainder when $g(x)=4x^3+4x^2-9x+5$ is divided by $(2x-1)$.

Mark schemeshow ▾
M1 root of divisor: $x=\tfrac12$  M1 $g\!\left(\tfrac12\right)=\tfrac12+1-\tfrac92+5$  A2 $=2$.
Q4. 5 marks

$(x-3)$ is a factor of $f(x)=x^3+ax^2-x-15$. (a) Show that $a=-1$. (b) Factorise $f(x)$ as far as possible, and explain why it cannot be written as a product of three linear factors.

Mark schemeshow ▾
(a) M1 $f(3)=27+9a-3-15=0$  A1 $9a=-9$, $a=-1$ ∎.
(b) M1 $f(x)=(x-3)(x^2+2x+5)$  A1 quotient correct  B1 discriminant of quotient $=4-20<0$ → no further real factors.
Q5. 5 marks

$f(x)=2x^3+px^2+qx-6$. Given that $(x-1)$ is a factor of $f(x)$, and that the remainder when $f(x)$ is divided by $(x+2)$ is $-36$, find $p$ and $q$.

Mark schemeshow ▾
M1 $f(1)=2+p+q-6=0 \Rightarrow p+q=4$  M1 $f(-2)=-16+4p-2q-6=-36 \Rightarrow 2p-q=-7$  M1 solve simultaneously  A1 $p=-1$  A1 $q=5$.
Q6. 6 marks

(a) Factorise $2x^3+x^2-25x+12$ completely. (b) Hence solve $2x^3+x^2-25x+12=0$.

Mark schemeshow ▾
(a) M1 trial: $f(3)=54+9-75+12=0$  A1 $(x-3)$ is a factor  M1 $(x-3)(2x^2+7x-4)$  A1 $(x-3)(2x-1)(x+4)$.
(b) M1 set each bracket $=0$  A1 $x=3,\ \tfrac12,\ -4$.
Q7. 6 marks

$f(x)=x^3-3x^2-10x+24$. (a) Show that $(x-2)$ is a factor and factorise $f(x)$ completely. (b) State the coordinates of the points where the curve $y=f(x)$ crosses the coordinate axes.

Mark schemeshow ▾
(a) M1 $f(2)=8-12-20+24=0$, conclusion stated  M1 $(x-2)(x^2-x-12)$  A1 $(x-2)(x-4)(x+3)$.
(b) B1 $x$-axis: $(2,0),(4,0),(-3,0)$  B1 $y$-axis: $(0,24)$  B1 both fully correct.
Q8. 7 marks

(a) Divide $x^3+2x^2-5x+7$ by $(x-2)$, stating the quotient and remainder. (b) Hence write $x^3+2x^2-5x+7$ in the form $(x-2)Q(x)+R$, and verify your remainder using the remainder theorem.

Mark schemeshow ▾
(a) M1 long division / comparing coefficients  A1 quotient $x^2+4x+3$  A1 remainder $13$.
(b) M1 $x^3+2x^2-5x+7=(x-2)(x^2+4x+3)+13$  A1 identity stated  M1 $f(2)=8+8-10+7$  A1 $=13$ ✓ matches.
Q9. 7 marks

$(2x+1)$ and $(x-2)$ are both factors of $f(x)=2x^3+ax^2+bx-2$. Find $a$ and $b$, and hence factorise $f(x)$ completely.

Mark schemeshow ▾
M1 $f\!\left(-\tfrac12\right)=0 \Rightarrow a-2b=9$  M1 $f(2)=0 \Rightarrow 2a+b=-7$  M1 solve  A1 $a=-1$  A1 $b=-5$  M1 third factor by division  A1 $f(x)=(x-2)(2x+1)(x+1)$.
Q10. 8 marks

The polynomial $p(x)=x^3+kx+20$ has $(x+2)$ as a factor. (a) Find $k$. (b) Factorise $p(x)$ as a product of a linear and a quadratic factor. (c) Prove that $x=-2$ is the only real root of $p(x)=0$.

Mark schemeshow ▾
(a) M1 $p(-2)=-8-2k+20=0$  A1 $k=6$.
(b) M1 divide by $(x+2)$  A1 $p(x)=(x+2)(x^2-2x+10)$.
(c) M1 discriminant of $x^2-2x+10$: $4-40$  A1 $=-36<0$  A1 quadratic has no real roots  B1 so $x=-2$ is the only real root ∎.