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2.12 Polynomial functionsIB Maths: Analysis and Approaches HL: Subtopic test

10 questions, 27 marks

IB Maths: Analysis and Approaches HL

2.12 Polynomial functions

Total 27 marks

Name

Class

Date

  1. 1
    The polynomial p(x)=2x3+ax2+bx−6p(x) = 2x^{3} + ax^{2} + bx - 6, where a,b∈Ra, b \in \mathbb{R}, has a factor (x−2)(x - 2). When p(x)p(x) is divided by (x+1)(x + 1) the remainder is −6-6.
    (a)
    Which equation follows from the fact that (x−2)(x - 2) is a factor of p(x)p(x)?
    [1 mark]
    • A4a−2b=224a - 2b = 22
    • B4a+2b=104a + 2b = 10
    • C4a+2b=−104a + 2b = -10
    • D4a+2b=−164a + 2b = -16
    (b)
    Which equation follows from the remainder theorem?
    [1 mark]
    • Aa−b=2a - b = 2
    • Ba−b=14a - b = 14
    • Ca+b=−2a + b = -2
    • Da−b=−4a - b = -4
    (c)
    Find the value of aa and the value of bb.
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    The equation 3x4−6x3+kx2+5x−12=03x^{4} - 6x^{3} + kx^{2} + 5x - 12 = 0, where k∈Rk \in \mathbb{R}, has roots α\alpha, β\beta, γ\gamma and δ\delta.
    (a)
    Find α+β+γ+δ\alpha + \beta + \gamma + \delta.
    [1 mark]
    • A−2-2
    • B22
    • C66
    • D12\frac{1}{2}
    (b)
    Find αβγδ\alpha\beta\gamma\delta.
    [1 mark]
    • A−14-\frac{1}{4}
    • B44
    • C−12-12
    • D−4-4
    (c)
    Find the sum and the product of the numbers 2α2\alpha, 2β2\beta, 2γ2\gamma and 2δ2\delta.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    A cubic function ff has zeros at x=−2x = -2, x=1x = 1 and x=3x = 3, and the graph of y=f(x)y = f(x) meets the yy-axis at (0,12)(0, 12). The function gg is defined by g(x)=f(x)+pxg(x) = f(x) + px, where p∈Rp \in \mathbb{R}.
    (a)
    Find f(x)f(x), giving your answer in the form ax3+bx2+cx+dax^{3} + bx^{2} + cx + d.
    [3 marks]
    (b)
    Given that (x−2)(x - 2) is a factor of g(x)g(x), find the value of pp and find the other two zeros of gg.
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    A company makes closed boxes in the shape of cuboids. Box P has edges of length xx cm, (x+2)(x + 2) cm and (x+5)(x + 5) cm, and a volume of 120 cm3120\ \text{cm}^{3}. For box Q, the lengths in cm of its three different edges are the roots of the equation 2x3−27x2+118x−168=02x^{3} - 27x^{2} + 118x - 168 = 0.
    (a)
    Show that x3+7x2+10x−120=0x^{3} + 7x^{2} + 10x - 120 = 0. Hence use the factor theorem to find the dimensions of box P, justifying that there is only one possible box.
    [6 marks]
    (b)
    Without solving the equation, find the volume of box Q and the total length of its twelve edges. Given that one edge of box Q is 4 cm long, find the lengths of the other two edges.
    [6 marks]

    Total for question 4: 12 marks

End of questions