Multiply Polynomials: Questions and Step-by-Step Solutions

This page is designed to help Grade 9 students master the multiplication of polynomials. Multiplying polynomials requires a solid understanding of the distributive property (often remembered as FOIL for binomials) and the exponent rules.

Below, you will find practice questions ranging from simple monomial multiplication to expanding complex binomials and trinomials. Try to solve each problem on your own, then open the step-by-step solutions to check your work and understand the reasoning behind each algebraic step.

Practice Questions

  1. 1. Multiply the following polynomials

    1. \( (-2)(-2x) \)
    2. \( (x)(x + 1) \)
    3. \( -x^2(-x + 1) \)
    4. \( (-4x^3 - x)(2x - 1) \)
    5. \( (-4x^3 - y)(2x - y) \)
    6. \( (-7x^2 - 2x + 3)(2x^2 - x + 2) \)
    7. \( \left(-\dfrac{1}{3}x^2 + 4\right)\left(-\dfrac{1}{2}x + 9\right) \)
    View Step-by-Step Solutions

    1. Multiply the constants, then attach the variable:

      \[ (-2)(-2x) = 4x \]

    2. Distribute \(x\) to both terms inside the parenthesis:

      \[ \begin{aligned} (x)(x + 1) &= x(x) + x(1) \\ &= x^2 + x \end{aligned} \]

    3. Distribute \(-x^2\):

      \[ \begin{aligned} -x^2(-x + 1) &= -x^2(-x) - x^2(1) \\ &= x^3 - x^2 \end{aligned} \]

    4. Use the distributive property (FOIL method) for the two binomials:

      \[ \begin{aligned} (-4x^3 - x)(2x - 1) &= -4x^3(2x - 1) - x(2x - 1) \\ &= -4x^3(2x) - 4x^3(-1) - x(2x) - x(-1) \\ &= -8x^4 + 4x^3 - 2x^2 + x \end{aligned} \]

    5. Distribute each term of the first binomial into the second:

      \[ \begin{aligned} (-4x^3 - y)(2x - y) &= -4x^3(2x - y) - y(2x - y) \\ &= -4x^3(2x) - 4x^3(-y) - y(2x) - y(-y) \\ &= -8x^4 + 4x^3y - 2xy + y^2 \end{aligned} \]

    6. Distribute each of the three terms in the first trinomial into the second trinomial, then combine like terms:

      \[ \begin{aligned} &(-7x^2 - 2x + 3)(2x^2 - x + 2) \\ &= -7x^2(2x^2 - x + 2) - 2x(2x^2 - x + 2) + 3(2x^2 - x + 2) \\ &= (-14x^4 + 7x^3 - 14x^2) + (-4x^3 + 2x^2 - 4x) + (6x^2 - 3x + 6) \\ &= -14x^4 + (7x^3 - 4x^3) + (-14x^2 + 2x^2 + 6x^2) + (-4x - 3x) + 6 \\ &= -14x^4 + 3x^3 - 6x^2 - 7x + 6 \end{aligned} \]

    7. Distribute the fractions carefully:

      \[ \begin{aligned} &\left(-\dfrac{1}{3}x^2 + 4\right)\left(-\dfrac{1}{2}x + 9\right) \\ &= -\dfrac{1}{3}x^2\left(-\dfrac{1}{2}x\right) - \dfrac{1}{3}x^2(9) + 4\left(-\dfrac{1}{2}x\right) + 4(9) \\ &= \dfrac{1}{6}x^3 - 3x^2 - 2x + 36 \end{aligned} \]

  2. 2. Multiply and Add/Subtract to write as a single polynomial

    Remember to follow the order of operations: multiply the polynomials first, then add or subtract like terms.

    1. \( (2x - 1)(3x - 2) + 3x - 9 \)
    2. \( -(2x + 2) - (2x - 1)(x - 3) \)
    3. \( (-3x - 2)(y - 3) + (x - 5)(y - 6) \)
    View Step-by-Step Solutions

    1. First, expand the binomials, then combine the remaining terms:

      \[ \begin{aligned} (2x - 1)(3x - 2) + 3x - 9 &= [2x(3x - 2) - 1(3x - 2)] + 3x - 9 \\ &= [6x^2 - 4x - 3x + 2] + 3x - 9 \\ &= 6x^2 - 7x + 2 + 3x - 9 \\ &= 6x^2 - 4x - 7 \end{aligned} \]

    2. Distribute the negative signs and multiply the binomials. Use brackets to avoid sign errors:

      \[ \begin{aligned} -(2x + 2) - (2x - 1)(x - 3) &= -2x - 2 - [2x(x - 3) - 1(x - 3)] \\ &= -2x - 2 - [2x^2 - 6x - x + 3] \\ &= -2x - 2 - [2x^2 - 7x + 3] \\ &= -2x - 2 - 2x^2 + 7x - 3 \\ &= -2x^2 + 5x - 5 \end{aligned} \]

    3. Expand both sets of binomials separately, then combine like terms:

      \[ \begin{aligned} &(-3x - 2)(y - 3) + (x - 5)(y - 6) \\ &= [-3x(y - 3) - 2(y - 3)] + [x(y - 6) - 5(y - 6)] \\ &= [-3xy + 9x - 2y + 6] + [xy - 6x - 5y + 30] \\ &= (-3xy + xy) + (9x - 6x) + (-2y - 5y) + (6 + 30) \\ &= -2xy + 3x - 7y + 36 \end{aligned} \]

  3. 3. Use multiplication of polynomials to expand the following

    1. \( (x + 3y)^2 \)
    2. \( (2x - y)^2 \)
    3. \( (x - y)(x + y) \)
    4. \( (x - 3)^3 \)
    View Step-by-Step Solutions

    1. Write the square as multiplication of two identical binomials (or use the perfect square formula):

      \[ \begin{aligned} (x + 3y)^2 &= (x + 3y)(x + 3y) \\ &= x(x + 3y) + 3y(x + 3y) \\ &= x^2 + 3xy + 3xy + 9y^2 \\ &= x^2 + 6xy + 9y^2 \end{aligned} \]

    2. Expand the perfect square binomial:

      \[ \begin{aligned} (2x - y)^2 &= (2x - y)(2x - y) \\ &= 2x(2x - y) - y(2x - y) \\ &= 4x^2 - 2xy - 2yx + y^2 \\ &= 4x^2 - 4xy + y^2 \end{aligned} \]

    3. This is a difference of squares:

      \[ \begin{aligned} (x - y)(x + y) &= x(x + y) - y(x + y) \\ &= x^2 + xy - yx - y^2 \\ &= x^2 - y^2 \end{aligned} \]

    4. For a cube, multiply three times. Group two of them first:

      \[ \begin{aligned} (x - 3)^3 &= (x - 3)(x - 3)(x - 3) \\ &= (x - 3)[x(x - 3) - 3(x - 3)] \\ &= (x - 3)[x^2 - 3x - 3x + 9] \\ &= (x - 3)(x^2 - 6x + 9) \end{aligned} \]

      Now, multiply the binomial by the resulting trinomial:

      \[ \begin{aligned} (x - 3)(x^2 - 6x + 9) &= x(x^2 - 6x + 9) - 3(x^2 - 6x + 9) \\ &= (x^3 - 6x^2 + 9x) + (-3x^2 + 18x - 27) \\ &= x^3 - 9x^2 + 27x - 27 \end{aligned} \]

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