Unit 6 · Checkpoint quiz

Unit 6 checkpoint: Factoring

14 questions · about 30-35 minutes · no time limit

Answer every question, then press Grade my answers at the bottom. You'll see what you got right, a worked solution for each question, and links to the lessons worth revisiting. Pencil and paper are encouraged.

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Factor out the GCF: \(6x^3y - 9x^2y^2 + 3x^2y\).

Numbers: the GCF of 6, 9, and 3 is 3. Variables: every term has at least \(x^2\) and \(y\). So the GCF is \(3x^2y\).

Divide each term by \(3x^2y\): \(\dfrac{6x^3y}{3x^2y} = 2x\), \(\dfrac{-9x^2y^2}{3x^2y} = -3y\), \(\dfrac{3x^2y}{3x^2y} = 1\).

Answer: \(3x^2y(2x - 3y + 1)\). (The \(+1\) is needed: the last term did not disappear.)

Factor by grouping: \(x^3 - 4x^2 + 3x - 12\).

Group in pairs: \((x^3 - 4x^2) + (3x - 12)\).

Take the GCF out of each pair: \(x^2(x - 4) + 3(x - 4)\).

Both pieces share \((x - 4)\), so factor it out. Answer: \((x - 4)(x^2 + 3)\).

Factor \(x^2 - 2x - 35\).

Find two numbers that multiply to \(-35\) and add to \(-2\).

\(-7 \cdot 5 = -35\) and \(-7 + 5 = -2\).

Answer: \((x - 7)(x + 5)\). Check: \(x^2 + 5x - 7x - 35 = x^2 - 2x - 35\).

Factor \(x^2 - 11xy + 24y^2\).

Find two numbers that multiply to \(24\) and add to \(-11\): \(-3\) and \(-8\).

Each second term gets a \(y\), because the last term is \(24y^2\).

Answer: \((x - 3y)(x - 8y)\). Check: \(x^2 - 8xy - 3xy + 24y^2 = x^2 - 11xy + 24y^2\).

Factor \(6x^2 + 7x - 3\).

ac method: \(a \cdot c = 6 \cdot (-3) = -18\). Two numbers that multiply to \(-18\) and add to \(7\): \(9\) and \(-2\).

Split the middle term: \(6x^2 + 9x - 2x - 3\).

Group: \(3x(2x + 3) - 1(2x + 3) = (2x + 3)(3x - 1)\).

Answer: \((2x + 3)(3x - 1)\). Check: \(6x^2 - 2x + 9x - 3 = 6x^2 + 7x - 3\).

Factor completely: \(12x^2 - 22x - 20\).

GCF first: \(12x^2 - 22x - 20 = 2(6x^2 - 11x - 10)\).

For \(6x^2 - 11x - 10\): \(ac = 6 \cdot (-10) = -60\). Two numbers that multiply to \(-60\) and add to \(-11\): \(-15\) and \(4\).

Split and group: \(6x^2 - 15x + 4x - 10 = 3x(2x - 5) + 2(2x - 5) = (3x + 2)(2x - 5)\).

Answer: \(2(3x + 2)(2x - 5)\).

Factor \(16x^2 - 40x + 25\).

The first and last terms are perfect squares: \(16x^2 = (4x)^2\) and \(25 = 5^2\).

Check the middle term: \(2 \cdot 4x \cdot 5 = 40x\), and it is subtracted. So this is a perfect square trinomial \(a^2 - 2ab + b^2 = (a - b)^2\).

Answer: \((4x - 5)^2\).

Factor \(8x^3 - 27\).

This is a difference of cubes: \(8x^3 = (2x)^3\) and \(27 = 3^3\), so \(a = 2x\) and \(b = 3\).

Use \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\): \((2x - 3)\big((2x)^2 + (2x)(3) + 3^2\big)\).

Answer: \((2x - 3)(4x^2 + 6x + 9)\).

Which is the complete factorization of \(2x^3 - 18x\)?

  1. \(2x(x^2 - 9)\)
  2. \(2x(x - 3)^2\)
  3. \(x(2x - 6)(x + 3)\)
  4. \(2x(x - 3)(x + 3)\)

GCF first: \(2x^3 - 18x = 2x(x^2 - 9)\). That is equal, but \(x^2 - 9\) still factors, so (a) is not complete.

\(x^2 - 9\) is a difference of squares: \((x - 3)(x + 3)\). So the complete answer is \(2x(x - 3)(x + 3)\).

(b) is wrong: \((x - 3)^2 = x^2 - 6x + 9\). (c) is equal, but \(2x - 6\) still has a common factor of 2. Answer: (d).

Factor completely: \(3x^4 - 48\).

GCF first: \(3(x^4 - 16)\).

\(x^4 - 16 = (x^2)^2 - 4^2\) is a difference of squares: \((x^2 - 4)(x^2 + 4)\).

\(x^2 - 4\) is another difference of squares: \((x - 2)(x + 2)\). The sum \(x^2 + 4\) does not factor.

Answer: \(3(x^2 + 4)(x - 2)(x + 2)\).

Solve \(2x^2 + 5x = 12\). Give all solutions, separated by a comma.

Standard form first (one side equal to 0): \(2x^2 + 5x - 12 = 0\).

Factor: \(ac = -24\); use \(8\) and \(-3\): \(2x^2 + 8x - 3x - 12 = 2x(x + 4) - 3(x + 4) = (2x - 3)(x + 4)\).

Zero product property: \(2x - 3 = 0\) gives \(x = \dfrac{3}{2}\); \(x + 4 = 0\) gives \(x = -4\).

Answer: \(x = \dfrac{3}{2}\) or \(x = -4\).

A rectangle's length is 3 meters more than its width. Its area is 40 square meters. What is the width, in meters?

Let \(w\) = width, so the length is \(w + 3\). Area: \(w(w + 3) = 40\).

Standard form: \(w^2 + 3w - 40 = 0\). Factor: \((w + 8)(w - 5) = 0\), so \(w = -8\) or \(w = 5\).

A width can't be negative, so \(w = 5\). Check: \(5 \cdot 8 = 40\). Answer: 5 meters.

Which is equal to \((x - 6)^2\)?

  1. \(x^2 - 12x - 36\)
  2. \(x^2 - 12x + 36\)
  3. \(x^2 + 12x + 36\)
  4. \(x^2 - 6x + 36\)

Use \((a - b)^2 = a^2 - 2ab + b^2\) with \(a = x\) and \(b = 6\).

\(x^2 - 2(x)(6) + 6^2 = x^2 - 12x + 36\).

The last term is \((-6)^2 = +36\). Answer: (b).

Solve \(\dfrac{x}{3} + \dfrac{x - 1}{4} = 5\).

The LCD of 3 and 4 is 12. Multiply every term by 12: \(4x + 3(x - 1) = 60\).

Distribute and combine: \(4x + 3x - 3 = 60\), so \(7x = 63\).

\(x = 9\). Check: \(\dfrac{9}{3} + \dfrac{8}{4} = 3 + 2 = 5\). Answer: \(x = 9\).

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