Factoring Polynomials: practice solutions, Grade 9 – download the PDF

1 Finding the GCF ★★★

Coefficients: GCF of 12 and 18 is 6. Variable \(a\): smallest exponent is 1. Variable \(b\): smallest exponent is 1.

The GCF is \(6ab\). Check: \(12a^2b = 6ab \cdot 2a\) and \(18ab^3 = 6ab \cdot 3b^2\).

2 Factoring out a GCF ★★★

a) GCF 5: \(5(3x + 5)\).

b) GCF \(4y\): \(4y(2y - 3)\).

c) GCF \(3m^2\): \(3m^2(3m + 2)\).

3 Easy trinomials ★★★

a) Product 10, sum 7: 2 and 5, so \((x + 2)(x + 5)\).

b) Product 21, sum 10: 3 and 7, so \((x + 3)(x + 7)\).

c) Product 24, sum 11: 3 and 8, so \((x + 3)(x + 8)\).

4 Trinomials with negative terms ★★★

a) Product \(-12\), sum \(-1\): \(-4\) and 3, so \((x - 4)(x + 3)\).

b) Product \(-15\), sum 2: 5 and \(-3\), so \((x + 5)(x - 3)\).

c) Product 14, sum \(-9\): \(-2\) and \(-7\), so \((x - 2)(x - 7)\).

5 Differences of squares ★★★

a) \(x^2 - 9^2 = (x - 9)(x + 9)\).

b) \((2y)^2 - 5^2 = (2y - 5)(2y + 5)\).

c) \(10^2 - n^2 = (10 - n)(10 + n)\).

6 Reading factored equations ★★★

a) \(x - 4 = 0\) or \(x + 9 = 0\): \(x = 4\) or \(x = -9\).

b) \(x = 0\) or \(2x - 7 = 0\): \(x = 0\) or \(x = \dfrac{7}{2}\).

c) \(3x + 6 = 0\) gives \(x = -2\); \(x - 1 = 0\) gives \(x = 1\).

7 True or false? ★★★

a) False. \((x + 4)(x - 4) = x^2 - 16\), not \(x^2 + 16\); a sum of squares does not factor this way.

b) True. \(3(2x + 3) = 6x + 9\).

c) True. \((x + 5)^2 = x^2 + 10x + 25\).

8 Factor completely ★★★

The GCF is \(6x\): \(12x^3 - 18x^2 + 30x = 6x(2x^2 - 3x + 5)\).

The trinomial \(2x^2 - 3x + 5\) has \(ac = 10\) and no pair of integers with product 10 and sum \(-3\), so it is prime. The complete factorization is \(6x(2x^2 - 3x + 5)\).

9 Grouping practice ★★★

a) \(x^2(x + 5) + 2(x + 5) = (x + 5)(x^2 + 2)\).

b) \(3a(b + 2) - 4(b + 2) = (b + 2)(3a - 4)\). Expanding: \(3ab - 4b + 6a - 8\), which matches.

10 Larger constants ★★★

a) Product \(-54\), sum \(-3\): \(-9\) and 6, so \((x - 9)(x + 6)\).

b) Product \(-56\), sum 1: 8 and \(-7\), so \((x + 8)(x - 7)\).

11 The ac method ★★★

a) \(ac = 6\), sum 7: 6 and 1. \(2x^2 + 6x + x + 3 = 2x(x + 3) + (x + 3) = (x + 3)(2x + 1)\).

b) \(ac = 30\), sum \(-17\): \(-15\) and \(-2\). \(5x^2 - 15x - 2x + 6 = 5x(x - 3) - 2(x - 3) = (x - 3)(5x - 2)\).

12 GCF and squares together ★★★

a) \((3x)^2 - 7^2 = (3x - 7)(3x + 7)\).

b) Factor out 2 first: \(2(25x^2 - 1) = 2(5x - 1)(5x + 1)\).

13 Perfect square trinomials ★★★

a) \(x^2 - 2\cdot 6x + 6^2 = (x - 6)^2\).

b) Middle term: \(2 \cdot 7x \cdot 2 = 28x\), so \((7x + 2)^2\).

c) Middle term: \(2 \cdot 3y \cdot 4 = 24y\), so \((3y - 4)^2\).

14 First equations ★★★

a) \((x - 3)(x - 5) = 0\), so \(x = 3\) or \(x = 5\).

b) Move 28: \(x^2 + 3x - 28 = 0\), so \((x + 7)(x - 4) = 0\) and \(x = -7\) or \(x = 4\). Check \(x = 4\): \(16 + 12 = 28\). Check \(x = -7\): \(49 - 21 = 28\).

15 Reading a graph ★★★

a) The graph crosses the x-axis at \(x = -2\) and \(x = 4\).

b) Product \(-8\), sum \(-2\): \(-4\) and 2, so \(x^2 - 2x - 8 = (x - 4)(x + 2)\), which is zero at \(x = 4\) and \(x = -2\).

c) At \(x = 0\), \(y = -8\); this equals the constant term.

16 Garden dimensions ★★★

a) Product 14, sum 9: 2 and 7, so area \(= (x + 2)(x + 7)\). The dimensions are \(x + 2\) m and \(x + 7\) m.

b) For \(x = 3\): width 5 m, length 10 m. Check the area: \(5 \cdot 10 = 50\) and \(9 + 27 + 14 = 50\). Perimeter \(= 2(5 + 10) = 30\) m (about 98.4 ft).

17 A trinomial equation ★★★

\(ac = -12\), sum 1: 4 and \(-3\). \(6x^2 + 4x - 3x - 2 = 2x(3x + 2) - (3x + 2) = (3x + 2)(2x - 1)\).

Then \(x = -\dfrac{2}{3}\) or \(x = \dfrac{1}{2}\).

18 Squares and common factors ★★★

a) \(4x^2 - 25 = 0\), \((2x - 5)(2x + 5) = 0\), so \(x = \pm\dfrac{5}{2}\).

b) \(3x^2 - 12x = 0\), \(3x(x - 4) = 0\), so \(x = 0\) or \(x = 4\). Dividing by \(x\) would have lost \(x = 0\).

19 Consecutive integers ★★★

Let the integers be \(n\) and \(n + 1\). Then \(n(n + 1) = 156\), so \(n^2 + n - 156 = 0\).

Product \(-156\), sum 1: 13 and \(-12\). \((n + 13)(n - 12) = 0\), so \(n = -13\) or \(n = 12\).

Only \(n = 12\) is positive. The integers are 12 and 13, and \(12 \cdot 13 = 156\).

20 A launched ball ★★★

Set \(-16t^2 + 64t + 80 = 0\). Factor out \(-16\): \(-16(t^2 - 4t - 5) = 0\), so \(-16(t - 5)(t + 1) = 0\).

Thus \(t = 5\) or \(t = -1\). A negative time makes no sense here, so the ball lands after 5 seconds. Check: \(-16 \cdot 25 + 64 \cdot 5 + 80 = -400 + 320 + 80 = 0\).

21 Find the error ★★★

The zero product property works only when the product equals 0, not 14.

Expand and set to zero: \(x^2 + x - 6 = 14\), so \(x^2 + x - 20 = 0\), \((x + 5)(x - 4) = 0\). Hence \(x = -5\) or \(x = 4\).

Check: \(x = 4\): \(2 \cdot 7 = 14\). \(x = -5\): \((-7)(-2) = 14\).

22 Which values of k? ★★★

We need integers \(p\) and \(q\) with \(pq = 12\); then \(k = p + q\).

Pairs: (1, 12), (2, 6), (3, 4) give sums 13, 8, 7. Pairs (\(-1\), \(-12\)), (\(-2\), \(-6\)), (\(-3\), \(-4\)) give sums \(-13\), \(-8\), \(-7\).

So \(k \in \{-13, -8, -7, 7, 8, 13\}\).