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Factoring of Quadratics - Solutions to Set 1

Using the factoring method to solve for the following quadratics expression.

Problem 1

Problem Statement: \(6x^2 - 11x - 35\)

Steps:

  1. Identify \(a = 6\), \(b = -11\), and \(c = -35\).
  2. Find two numbers that multiply to \(a \cdot c = 6 \cdot (-35) = -210\) and add to \(b = -11\). These numbers are \(-15\) and \(14\) because \(-15 \cdot 14 = -210\) and \(-15 + 14 = -11\).
  3. Rewrite the middle term using these numbers:
\[ 6x^2 - 15x + 14x - 35 \]
  1. Factor by grouping:
\[ (6x^2 - 15x) + (14x - 35) \]
\[ 3x(2x - 5) + 7(2x - 5) \]
  1. Factor out the common binomial factor:
\[ (3x + 7)(2x - 5) \]

Problem 2.

Problem Statement: \(x^2 + 10x + 24\)

Steps:

  1. Identify \(a = 1\), \(b = 10\), and \(c = 24\).
  2. Find two numbers that multiply to \(a \cdot c = 24\) and add to \(b = 10\). These numbers are \(4\) and \(6\) because \(4 \cdot 6 = 24\) and \(4 + 6 = 10\).
  3. Rewrite the middle term using these numbers:
\[ x^2 + 4x + 6x + 24 \]
  1. Factor by grouping: [ (x^2 + 4x) + (6x + 24) ]

[ x(x + 4) + 6(x + 4) ] 5. Factor out the common binomial factor:

\[ (x + 4)(x + 6) \]

Problem 3.

Problem Statement: \(\frac{1}{2}x^2 - \frac{5}{2}x + 3\)

Steps:

  1. Identify \(a = \frac{1}{2}\), \(b = -\frac{5}{2}\), and \(c = 3\).
  2. Multiply \(a\) and \(c\):
\[ \frac{1}{2} \cdot 3 = \frac{3}{2} \]
  1. Find two numbers that multiply to \(\frac{3}{2}\) and add to \(-\frac{5}{2}\). These numbers are \(-1\) and \(-\frac{3}{2}\) because \(-1 \cdot -\frac{3}{2} = \frac{3}{2}\) and \(-1 - \frac{3}{2} = -\frac{5}{2}\).
  2. Rewrite the middle term using these numbers:
\[ \frac{1}{2}x^2 - x - \frac{3}{2}x + 3 \]
  1. Factor by grouping:
\[ \frac{1}{2}x(x - 2) - \frac{3}{2}(x - 2) \]
  1. Factor out the common binomial factor:
\[ \frac{1}{2}(x - 2)(x - 3) \]

Problem 4.

Problem Statement: \(4x^2 - 25\) Steps:

  1. Recognize this as a difference of squares:
\[ 4x^2 - 25 = (2x)^2 - 5^2 \]
  1. Apply the difference of squares formula \(a^2 - b^2 = (a - b)(a + b)\): [ (2x - 5)(2x + 5) ]

Problem 5.

Problem Statement: \(2x^2 + 7x - 15\)

Steps:

  1. Identify \(a = 2\), \(b = 7\), and \(c = -15\).
  2. Find two numbers that multiply to \(a \cdot c = 2 \cdot (-15) = -30\) and add to \(b = 7\). These numbers are \(10\) and \(-3\) because \(10 \cdot -3 = -30\) and \(10 - 3 = 7\).
  3. Rewrite the middle term using these numbers:
\[ 2x^2 + 10x - 3x - 15 \]
  1. Factor by grouping:
\[ (2x^2 + 10x) + (-3x - 15) \]
\[ 2x(x + 5) - 3(x + 5) \]
  1. Factor out the common binomial factor:
\[ (2x - 3)(x + 5) \]

Problem 6.

Problem Statement: \(3x^2 - 12x - 45\)

Steps:

  1. Factor out the common factor of \(3\): [ 3(x^2 - 4x - 15) ]
  2. Identify \(a = 1\), \(b = -4\), and \(c = -15\).
  3. Find two numbers that multiply to \(a \cdot c = -15\) and add to \(b = -4\). These numbers are \(-5\) and \(3\) because \(-5 \cdot 3 = -15\) and \(-5 + 3 = -4\).
  4. Rewrite the middle term using these numbers:
\[ x^2 - 5x + 3x - 15 \]
  1. Factor by grouping:
\[ (x^2 - 5x) + (3x - 15) \]
\[ x(x - 5) + 3(x - 5) \]
  1. Factor out the common binomial factor:
\[ (x - 5)(x + 3) \]
  1. Multiply by the common factor:
\[ 3(x - 5)(x + 3) \]

Problem 7.

Problem Statement: \(5x^2 + 23x + 12\)

Steps:

  1. Identify \(a = 5\), \(b = 23\), and \(c = 12\).
  2. Find two numbers that multiply to \(a \cdot c = 5 \cdot 12 = 60\) and add to \(b = 23\). These numbers are \(15\) and \(4\) because \(15 \cdot 4 = 60\) and \(15 + 4 = 23\).
  3. Rewrite the middle term using these numbers:
\[ 5x^2 + 15x + 4x + 12 \]
  1. Factor by grouping:
\[ (5x^2 + 15x) + (4x + 12) \]
\[ 5x(x + 3) + 4(x + 3) \]
  1. Factor out the common binomial factor: [ (5x + 4)(x + 3) ]

Problem 8.

Problem Statement: \(-3x^2 + 8x + 16\)

Steps:

  1. Factor out the common factor of \(-1\): [ -1(3x^2 - 8x - 16) ]

  2. Identify \(a = 3\), \(b = -8\), and \(c = -16\).

  3. Find two numbers that multiply to \(a \cdot c = 3 \cdot (-16) = -48\) and add to \(b = -8\). These numbers are \(-12\) and \(4\) because \(-12 \cdot 4 = -48\) and \(-12 + 4 = -8\).
  4. Rewrite the middle term using these numbers:
\[ -1(3x^2 - 12x + 4x - 16) \]
  1. Factor by grouping: [ -1[(3x^2 - 12x) + (4x - 16)] ]
\[ -1[3x(x - 4) + 4(x - 4)] \]
  1. Factor out the common binomial factor:
\[ -1(x - 4)(3x + 4) \]

Problem 9.

Problem Statement: \(x^2 - 10x + 25\)

Steps:

  1. Recognize this as a perfect square trinomial:
\[ x^2 - 10x + 25 = (x - 5)^2 \]

Problem 10.

Problem Statement: \(x^2 - 6x + 8\)

Steps: 1. Identify \(a = 1\), \(b = -6\), and \(c = 8\). 2. Find two numbers that multiply to \(a \cdot c = 8\) and add to \(b = -6\). These numbers are \(-2\) and \(-4\) because \(-2 \cdot -4 = 8\) and \(-2 - 4 = -6\). 3. Rewrite the middle term using these numbers:

\[ x^2 - 2x - 4x + 8 \]
  1. Factor by grouping:
\[ (x^2 - 2x) + (-4x + 8) \]
\[ x(x - 2) - 4(x - 2) \]
  1. Factor out the common binomial factor:
\[ (x - 2)(x - 4) \]

Problem 11.

Problem Statement: \(x^2 + 4x + 8\)

Steps:

  1. Identify \(a = 1\), \(b = 4\), and \(c = 8\).
  2. Find two numbers that multiply to \(a \cdot c = 8\) and add to \(b = 4\). This quadratic does not factor nicely over the reals.
  3. Use the quadratic formula to find roots:
\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} \]
\[ x = \frac{-4 \pm \sqrt{16 - 32}}{2} \]
\[ x = \frac{-4 \pm \sqrt{-16}}{2} \]
\[ x = \frac{-4 \pm 4i}{2} \]
\[ x = -2 \pm 2i \]

Thus, the factors are:

\[ (x - (-2 + 2i))(x - (-2 - 2i)) = (x + 2 - 2i)(x + 2 + 2i) \]

Problem 12.

Problem Statement: \(6x^2 + 11x - 35\)

Steps:

  1. Identify \(a = 6\), \(b = 11\), and \(c = -35\).
  2. Find two numbers that multiply to \(a \cdot c = 6 \cdot (-35) = -210\) and add to \(b = 11\). These numbers are \(15\) and \(-14\) because \(15 \cdot -14 = -210\) and \(15 - 14 = 11\).
  3. Rewrite the middle term using these numbers:
\[ 6x^2 + 15x - 4x - 35 \]
  1. Factor by grouping:
\[ (6x^2 + 15x) - (4x + 35) \]
\[ 3x(2x + 5) - 7(2x + 5) \]
  1. Factor out the common binomial factor:
\[ (3x - 7)(2x + 5) \]

These steps guide you through the process of factoring each quadratic expression. If you have any questions or need further clarification, feel free to ask!