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Neokaal Study

Polynomials

Practice Worksheet

Zeroes, coefficient relations, factorisation, polynomial construction, and graph interpretation.

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  1. Factorise $3x^2 + 27x + 60$, find its zeroes, and verify the coefficient relations.

  2. Find a quadratic polynomial with zeroes adding to $-\frac{1}{2}$ and multiplying to $-\frac{3}{2}$; also find the zeroes.

  3. Use factorisation to solve $5y^2 - 3\sqrt{5}y - 10$ = $0$, and verify the relations between zeroes and coefficients.

  4. Use the given sum $-\frac{7\sqrt{5}}{5}$ and product $\frac{12}{5}$ of zeroes to build a quadratic polynomial and factorise it.

  5. Choose the sketch that is not a parabola and hence not a quadratic graph. (A) (B) (C) (D)

    A
    downward-opening parabola touching the x-axis
    xy
    B
    downward-opening parabola crossing the x-axis twice
    xy
    C
    upward-opening parabola above the x-axis
    xy
    D
    S-shaped curve crossing the x-axis three times
    xy

  6. Write true or false, with a reason: If the graph of a quadratic polynomial intersects the x-axis at only one point, its two zeroes cannot be equal.

    upward-opening parabola touching the x-axis once
    xy

Neokaal Study

Answer Key

  1. The zeroes are $-4$ and $-5$.

    1. Factorise the polynomial: $3x^2 + 27x + 60$ = $3(x + 4)(x + 5)$.
    2. Set each factor equal to $0$; this gives $x = -4$ and $x = -5$.
    3. Sum of zeroes: $-9$; from coefficients, $-b/a = -9$.
    4. Product of zeroes: $20$; from coefficients, $c/a = 20$.

  2. One suitable polynomial is $y^2 + \frac{1}{2}y - \frac{3}{2}$; its zeroes are $1$ and $-\frac{3}{2}$.

    1. For zeroes with sum $-\frac{1}{2}$ and product $-\frac{3}{2}$, use $y^2 - (sum)y + product$.
    2. So the polynomial is $y^2 + \frac{1}{2}y - \frac{3}{2}$.
    3. Factorising gives $y^2 + \frac{1}{2}y - \frac{3}{2}$ = $(y - 1)(y + \frac{3}{2})$.
    4. Hence the zeroes are $1$ and $-\frac{3}{2}$.

  3. The zeroes are $\sqrt{5}$ and $-\frac{2\sqrt{5}}{5}$.

    1. Factorise the polynomial: $5y^2 - 3\sqrt{5}y - 10$ = $(\sqrt{5}y - 5)(\sqrt{5}y + 2)$.
    2. Set each factor equal to $0$; this gives $y = \sqrt{5}$ and $y = -\frac{2\sqrt{5}}{5}$.
    3. Sum of zeroes: $\frac{3\sqrt{5}}{5}$; from coefficients, $-b/a = \frac{3\sqrt{5}}{5}$.
    4. Product of zeroes: $-2$; from coefficients, $c/a = -2$.

  4. One suitable polynomial is $t^2 + \frac{7\sqrt{5}}{5}t + \frac{12}{5}$; its zeroes are $-\frac{3\sqrt{5}}{5}$ and $-\frac{4\sqrt{5}}{5}$.

    1. For zeroes with sum $-\frac{7\sqrt{5}}{5}$ and product $\frac{12}{5}$, use $t^2 - (sum)t + product$.
    2. So the polynomial is $t^2 + \frac{7\sqrt{5}}{5}t + \frac{12}{5}$.
    3. Factorising gives $t^2 + \frac{7\sqrt{5}}{5}t + \frac{12}{5}$ = $(t + \frac{3\sqrt{5}}{5})(t + \frac{4\sqrt{5}}{5})$.
    4. Hence the zeroes are $-\frac{3\sqrt{5}}{5}$ and $-\frac{4\sqrt{5}}{5}$.

  5. Option D is not a quadratic graph.

    1. A quadratic polynomial has a parabolic graph.
    2. A parabola can open upward or downward, and it may miss, touch, or cross the x-axis.
    3. The S-shaped sketch has cubic-type behaviour, so it is not a quadratic graph.
    4. Therefore the answer is option $D$.

  6. False. A quadratic graph that touches the x-axis has two equal zeroes.

    1. The zeroes of a polynomial are the x-coordinates where its graph meets the x-axis.
    2. A quadratic with equal zeroes has a parabola tangent to the x-axis.
    3. That graph has one x-intercept representing the repeated zero.
    4. Therefore one x-axis intersection is consistent with two equal zeroes.