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← Introduction to Algebra levels

Introduction to Algebra / LEVEL 2 · DIFFICULTY 2/5

Products, Roots and Turning Points

Connect factoring to solutions and graph structure.

3 stages · 24 practice problems · two 6-question assessment forms

STAGE ALG 2.1 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Expanding and Collecting Terms

Useful preparation: Slope and Rate of Change

Goal: Identify coefficients in a product of linear expressions.

Before you begin: Slope and Rate of Change

Understand the idea

Multiplication distributes over addition: every term in one factor multiplies every term in the other. Like terms have the same variable powers, so their coefficients can be added.

(x+a) (x−b)=x2+(a−b) x−ab\displaystyle \left(x + a\right)\, \left(x - b\right) = x^{2} + \left(a - b\right)\, x - ab

Choose and carry out a method

Make four products for two binomials, keep their signs, then collect x2\displaystyle x^{2} terms, x terms and constants separately. A rectangle model shows why each product appears exactly once.

Check the reasoning

Substitute a small allowed value into both the factored and expanded forms. Agreement is a useful check, although one test value alone does not prove an identity.

WORKED EXAMPLE 1

Find the coefficient of x in (x+3) (x−6)\displaystyle \left(x + 3\right)\, \left(x - 6\right).

  1. Distribute each term and combine like powers.
  2. (x+3) (x−6)=x2+(3−6) x−18\displaystyle \left(x + 3\right)\, \left(x - 6\right) = x^{2} + \left(3 - 6\right)\, x - 18.
  3. The coefficient of x is -3. The constant term is a different coefficient.

-3

WORKED EXAMPLE 2

Find the coefficient of x in (x+4) (x−7)\displaystyle \left(x + 4\right)\, \left(x - 7\right).

  1. Distribute each term and combine like powers.
  2. (x+4) (x−7)=x2+(4−7) x−28\displaystyle \left(x + 4\right)\, \left(x - 7\right) = x^{2} + \left(4 - 7\right)\, x - 28.
  3. The coefficient of x is -3. The constant term is a different coefficient.

-3

Common pitfalls

Possible mix-up: Multiply only the first and last terms.

The two cross-products also contribute.

Possible mix-up: An x term can combine with an x2\displaystyle x^{2} term.

Only equal variable powers are like terms.

Explain it to yourself

Explain where each of the four products appears in a split rectangle.

Preview the eight practice prompts
  1. Problem 1

    Find the coefficient of x in (x+6) (x−9)\displaystyle \left(x + 6\right)\, \left(x - 9\right).

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  2. Problem 2

    Find the coefficient of x in (x+7) (x−5)\displaystyle \left(x + 7\right)\, \left(x - 5\right).

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  3. Problem 3

    Find the coefficient of x in (x+8) (x−6)\displaystyle \left(x + 8\right)\, \left(x - 6\right).

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  4. Problem 4

    Find the coefficient of x in (x+9) (x−7)\displaystyle \left(x + 9\right)\, \left(x - 7\right).

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  5. Problem 5

    Find the coefficient of x in (x+10) (x−8)\displaystyle \left(x + 10\right)\, \left(x - 8\right).

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  6. Problem 6

    Find the coefficient of x in (x+11) (x−9)\displaystyle \left(x + 11\right)\, \left(x - 9\right).

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  7. Problem 7New context

    A rectangle has sides (x+12)\displaystyle \left(x + 12\right) and (x−5)\displaystyle \left(x - 5\right), with x>5\displaystyle x > 5. When its area is expanded, what is the coefficient of x?

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  8. Problem 8New context

    A rectangle has sides (x+13)\displaystyle \left(x + 13\right) and (x−6)\displaystyle \left(x - 6\right), with x>6\displaystyle x > 6. When its area is expanded, what is the coefficient of x?

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Open stage ALG 2.1 in the student workspace →

STAGE ALG 2.2 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Factoring Quadratics

Useful preparation: Expanding and Collecting Terms

Goal: Use a product and a sum to recover quadratic roots.

Before you begin: Expanding and Collecting Terms

AN EQUATION IS A RULE FOR MAKING POINTS

Change the rule. Watch the graph.

Pick an x value. The rule tells you its y value. A point (x, y) shows both numbers together. Drag across the graph or use the point slider. This is an adjustable learning example; use the coefficients in your own practice question when solving it.

Parent equation: y=x2\displaystyle y = x^{2}

Start with y=x2\displaystyle y = x^{2}. Complete the square: y=1 (x−(1))2+(−4)\displaystyle y = 1\, \left(x - \left(1\right)\right)^{2} + \left(-4\right). The new vertex is (1, -4). The first coordinate gives the horizontal shift and the second gives the vertical shift. Multiply vertical distances from the vertex by 1.

Trace a point along the graph
0%

Blue: y=1 x2−2 x−3\displaystyle y = 1\, x^{2} - 2\, x - 3

y = 1x² − 2x − 3; horizontal x-axis and vertical y-axis-10-10-8-8-6-6-4-4-2-2224466881010xy0

Moving point: (-4, 21)

Dashed gray: parent equation. Blue: current equation. View: −10 to 10 on both axes. The graph continues beyond this window. Purple marks your selected point.

Curve coefficient a: 1
Coefficient b: -2
Vertical intercept c: -3
Choose the point’s x: 1
Point (1, -4)
1×(1)2+(−2)×(1)+(−3)=−4\displaystyle 1 \times \left(1\right)^{2} + \left(-2\right) \times \left(1\right) + \left(-3\right) = -4
Why does it bend?

The x2\displaystyle x^{2} term makes the change in y vary with x. This parabola opens upward because a is positive.

Three points made by the blue rule
xSubstitute into the ruley
-11×(−1)2+(−2)×(−1)+(−3)\displaystyle 1 \times \left(-1\right)^{2} + \left(-2\right) \times \left(-1\right) + \left(-3\right)0
01×(0)2+(−2)×(0)+(−3)\displaystyle 1 \times \left(0\right)^{2} + \left(-2\right) \times \left(0\right) + \left(-3\right)-3
11×(1)2+(−2)×(1)+(−3)\displaystyle 1 \times \left(1\right)^{2} + \left(-2\right) \times \left(1\right) + \left(-3\right)-4

Understand the idea

Factoring rewrites a polynomial as a product. If a product equals zero, at least one factor is zero. This connects algebraic structure to the x-intercepts of a quadratic graph.

(x−r) (x−s)=x2−(r+s) x+rs\displaystyle \left(x - r\right)\, \left(x - s\right) = x^{2} - \left(r + s\right)\, x + rs

Choose and carry out a method

For a monic quadratic, seek two numbers whose sum is the negative of the x coefficient and whose product is the constant. Expand the proposed factors before using the zero-product rule.

Check the reasoning

Check both roots in the original polynomial. In a geometric model, reject lengths that are nonpositive, and distinguish the smaller root from the sum of roots.

WORKED EXAMPLE 1

The equation x2−9 x+18=0\displaystyle x^{2} - 9\, x + 18 = 0 has two roots. Enter the smaller root.

  1. Look for two numbers with the required product and sum.
  2. The numbers 3 and 6 multiply to 18 and add to 9.
  3. The equation factors as (x−3) (x−6)=0\displaystyle \left(x - 3\right)\, \left(x - 6\right) = 0, so the smaller root is 3.

3

WORKED EXAMPLE 2

The equation x2−11 x+28=0\displaystyle x^{2} - 11\, x + 28 = 0 has two roots. Enter the smaller root.

  1. Look for two numbers with the required product and sum.
  2. The numbers 4 and 7 multiply to 28 and add to 11.
  3. The equation factors as (x−4) (x−7)=0\displaystyle \left(x - 4\right)\, \left(x - 7\right) = 0, so the smaller root is 4.

4

Common pitfalls

Possible mix-up: A positive constant means both roots are negative.

The sum of the roots also determines their signs.

Possible mix-up: A product can be set to zero even if the equation equals 5.

First rewrite the equation with zero on one side.

Explain it to yourself

What feature of the equation lets you set each factor equal to zero?

Preview the eight practice prompts
  1. Problem 1

    The equation x2−15 x+54=0\displaystyle x^{2} - 15\, x + 54 = 0 has two roots. Enter the smaller root.

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  2. Problem 2

    The equation x2−17 x+70=0\displaystyle x^{2} - 17\, x + 70 = 0 has two roots. Enter the smaller root.

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  3. Problem 3

    The equation x2−19 x+88=0\displaystyle x^{2} - 19\, x + 88 = 0 has two roots. Enter the smaller root.

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  4. Problem 4

    The equation x2−21 x+108=0\displaystyle x^{2} - 21\, x + 108 = 0 has two roots. Enter the smaller root.

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  5. Problem 5

    The equation x2−23 x+130=0\displaystyle x^{2} - 23\, x + 130 = 0 has two roots. Enter the smaller root.

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  6. Problem 6

    The equation x2−25 x+154=0\displaystyle x^{2} - 25\, x + 154 = 0 has two roots. Enter the smaller root.

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  7. Problem 7New context

    A rectangle has area 180 square units and its side lengths add to 27 units. What is its shorter side length?

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  8. Problem 8New context

    A rectangle has area 208 square units and its side lengths add to 29 units. What is its shorter side length?

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Open stage ALG 2.2 in the student workspace →

STAGE ALG 2.3 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Completing the Square

Useful preparation: Factoring Quadratics

Goal: Find an attained quadratic minimum without guessing.

Before you begin: Factoring Quadratics

AN EQUATION IS A RULE FOR MAKING POINTS

Change the rule. Watch the graph.

Pick an x value. The rule tells you its y value. A point (x, y) shows both numbers together. Drag across the graph or use the point slider. This is an adjustable learning example; use the coefficients in your own practice question when solving it.

Parent equation: y=x2\displaystyle y = x^{2}

Start with y=x2\displaystyle y = x^{2}. Complete the square: y=1 (x−(1))2+(−4)\displaystyle y = 1\, \left(x - \left(1\right)\right)^{2} + \left(-4\right). The new vertex is (1, -4). The first coordinate gives the horizontal shift and the second gives the vertical shift. Multiply vertical distances from the vertex by 1.

Trace a point along the graph
0%

Blue: y=1 x2−2 x−3\displaystyle y = 1\, x^{2} - 2\, x - 3

y = 1x² − 2x − 3; horizontal x-axis and vertical y-axis-10-10-8-8-6-6-4-4-2-2224466881010xy0

Moving point: (-4, 21)

Dashed gray: parent equation. Blue: current equation. View: −10 to 10 on both axes. The graph continues beyond this window. Purple marks your selected point.

Curve coefficient a: 1
Coefficient b: -2
Vertical intercept c: -3
Choose the point’s x: 1
Point (1, -4)
1×(1)2+(−2)×(1)+(−3)=−4\displaystyle 1 \times \left(1\right)^{2} + \left(-2\right) \times \left(1\right) + \left(-3\right) = -4
Why does it bend?

The x2\displaystyle x^{2} term makes the change in y vary with x. This parabola opens upward because a is positive.

Three points made by the blue rule
xSubstitute into the ruley
-11×(−1)2+(−2)×(−1)+(−3)\displaystyle 1 \times \left(-1\right)^{2} + \left(-2\right) \times \left(-1\right) + \left(-3\right)0
01×(0)2+(−2)×(0)+(−3)\displaystyle 1 \times \left(0\right)^{2} + \left(-2\right) \times \left(0\right) + \left(-3\right)-3
11×(1)2+(−2)×(1)+(−3)\displaystyle 1 \times \left(1\right)^{2} + \left(-2\right) \times \left(1\right) + \left(-3\right)-4

Understand the idea

A square of a real number is nonnegative. Rewriting a quadratic as a square plus a constant reveals its turning point and the exact bound on its values.

x2−2\displaystyle x^{2} - 2hx+h2+k=(x−h)2+k\displaystyle h^{2} + k = \left(x - h\right)^{2} + k

Choose and carry out a method

Take half the x coefficient, square it, and add and subtract that same amount. The added terms change the appearance but not the value of the expression.

Check the reasoning

State where equality occurs. A lower bound is the minimum only when an allowed input reaches it. If the domain is restricted, the vertex may not be available.

WORKED EXAMPLE 1

Find the minimum value of x2−6 x+19\displaystyle x^{2} - 6\, x + 19 over all real x.

  1. Complete the square and use its nonnegativity.
  2. x2−6 x+19=(x−3)2+10\displaystyle x^{2} - 6\, x + 19 = \left(x - 3\right)^{2} + 10.
  3. The square can be zero at x=3\displaystyle x = 3, so the attained minimum is 10.

10

WORKED EXAMPLE 2

Find the minimum value of x2−8 x+29\displaystyle x^{2} - 8\, x + 29 over all real x.

  1. Complete the square and use its nonnegativity.
  2. x2−8 x+29=(x−4)2+13\displaystyle x^{2} - 8\, x + 29 = \left(x - 4\right)^{2} + 13.
  3. The square can be zero at x=4\displaystyle x = 4, so the attained minimum is 13.

13

Common pitfalls

Possible mix-up: Add a square without compensating.

Add and subtract the same square to preserve equality.

Possible mix-up: The minimum happens at x=0\displaystyle x = 0.

The squared term vanishes at x=h\displaystyle x = h.

Explain it to yourself

How would the answer change if x were required to lie to the right of the vertex?

Preview the eight practice prompts
  1. Problem 1

    Find the minimum value of x2−12 x+55\displaystyle x^{2} - 12\, x + 55 over all real x.

    Practise this exact problem →
  2. Problem 2

    Find the minimum value of x2−14 x+71\displaystyle x^{2} - 14\, x + 71 over all real x.

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  3. Problem 3

    Find the minimum value of x2−16 x+89\displaystyle x^{2} - 16\, x + 89 over all real x.

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  4. Problem 4

    Find the minimum value of x2−18 x+109\displaystyle x^{2} - 18\, x + 109 over all real x.

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  5. Problem 5

    Find the minimum value of x2−20 x+131\displaystyle x^{2} - 20\, x + 131 over all real x.

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  6. Problem 6

    Find the minimum value of x2−22 x+155\displaystyle x^{2} - 22\, x + 155 over all real x.

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  7. Problem 7New context

    The energy cost of a setting x is E (x)=x2−24 x+181\displaystyle E\, \left(x\right) = x^{2} - 24\, x + 181. Settings may be any real number. What is the lowest possible cost?

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  8. Problem 8New context

    The energy cost of a setting x is E (x)=x2−26 x+209\displaystyle E\, \left(x\right) = x^{2} - 26\, x + 209. Settings may be any real number. What is the lowest possible cost?

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Open stage ALG 2.3 in the student workspace →
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