Factor and complete the square to reveal the zeros and the maximum or minimum of a quadratic

Learning goal: Choose and produce an equivalent form of a quadratic expression to reveal and explain properties of the quantity it represents: factor a quadratic expression to reveal the zeros of the function it defines, and complete the square in a quadratic expression to reveal the maximum or minimum value of the function it defines.

Three forms of the same quadratic. A quadratic expression in x can be written in standard form ax² + bx + c, in factored form a(x − r)(x − s) when it factors over the numbers you are using, and in vertex form a(x − h)² + k. All three are equivalent: substituting any value of x gives the same number in each. They are not equally informative. Standard form displays the constant term c, the value at x = 0. Factored form displays the zeros r and s, the inputs that make the expression 0. Vertex form displays the maximum or minimum value k and where it occurs, x = h. The skill in this lesson is to decide which property is wanted and to produce the form that reveals it.

Factoring reveals zeros, because of the zero-product property. If a product of real numbers is 0, at least one factor is 0. So if x² − 5x − 14 is rewritten as (x − 7)(x + 2), the expression is 0 exactly when x − 7 = 0 or x + 2 = 0, that is at x = 7 and x = −2. Check both: 49 − 35 − 14 = 0 and 4 + 10 − 14 = 0. The factoring itself comes from searching for two numbers whose product is −14 and whose sum is −5: they are −7 and 2. Expand to confirm: (x − 7)(x + 2) = x² + 2x − 7x − 14 = x² − 5x − 14. The zero-product reasoning requires the expression to be equal to zero; it says nothing about (x − 7)(x + 2) = 6, where you cannot set each factor equal to 6.

When the leading coefficient is not 1. For 2x² + 7x − 4, look for factors (2x + p)(x + q) with pq = −4 and 2q + p = 7. Trying p = −1, q = 4: 2(4) + (−1) = 7, and (−1)(4) = −4. So 2x² + 7x − 4 = (2x − 1)(x + 4), with zeros x = 1/2 and x = −4. Expand to check: 2x² + 8x − x − 4 = 2x² + 7x − 4. In a context, zeros are events: a ball whose height is h(t) = −16t² + 64t (an idealized model that ignores air resistance and applies only during the flight, 0 ≤ t ≤ 4) factors as −16t(t − 4), so the height is 0 at t = 0 (launch) and t = 4 seconds (landing). The factored form reveals the landing time; the standard form does not show it.

Completing the square reveals the extreme value. A perfect-square trinomial has the pattern x² + 2mx + m² = (x + m)². Given x² − 6x + 11, the coefficient of x is −6, so m = −3 and the square that fits is (x − 3)² = x² − 6x + 9. Then x² − 6x + 11 = (x² − 6x + 9) + 2 = (x − 3)² + 2. The key reading: (x − 3)² is never negative and is 0 only at x = 3, so the expression is never less than 2 and equals 2 at x = 3. The minimum value is 2, reached at x = 3. Nothing in the standard form x² − 6x + 11 shows that; the vertex form makes it visible, and the argument is the same "a square is at least zero" reasoning used to interpret expressions by treating a part as a single entity.

A worked model inside the lesson, leading coefficient not 1. Find the maximum of −2x² + 12x − 5. Factor the leading coefficient out of the x-terms only: −2(x² − 6x) − 5. Inside, complete the square: x² − 6x = (x − 3)² − 9. Substitute and distribute the −2, which is the step students skip and get wrong: −2((x − 3)² − 9) − 5 = −2(x − 3)² + 18 − 5 = −2(x − 3)² + 13. Since −2(x − 3)² is never positive, the expression is never more than 13, and it equals 13 at x = 3. Maximum 13 at x = 3. Check by substituting x = 3 into the original: −18 + 36 − 5 = 13. Check a neighbor, x = 4: −32 + 48 − 5 = 11, which is less than 13, consistent with a maximum. A negative leading coefficient produces a maximum; a positive one produces a minimum.

Extreme values in context. Revenue R = p(120 − 4p) = −4p² + 120p, for prices 0 ≤ p ≤ 30 where the number sold, 120 − 4p, is not negative. Complete the square: −4(p² − 30p) = −4((p − 15)² − 225) = −4(p − 15)² + 900. Maximum revenue 900 dollars at price p = 15. The same information came from the factored form by symmetry between the zeros 0 and 30, and the two methods agreeing is a check. Vertex form goes further: it says how far revenue falls for a price away from 15, namely 4(p − 15)² dollars, so a price of 20 loses 4 × 25 = 100 dollars relative to the best.

Choosing the form. Ask what the question wants. "When is it zero?" or "when does it land?" or "what price gives no revenue?" calls for factored form. "What is the largest?" or "the least?" or "how high?" calls for completing the square. "What is the starting value?" is read from the standard form's constant term. Producing an equivalent form is only useful when it reveals the property you are asked to explain; producing all three every time is wasted work.

Misconceptions. Completing the square on 3x² + 12x + 7 by writing 3(x² + 4x + 4) + 7 − 4 is wrong, because the 4 added inside the parentheses was multiplied by 3: the correct compensation is 3(x + 2)² + 7 − 12 = 3(x + 2)² − 5, minimum −5 at x = −2. Check: 3(4) − 24 + 7 = −5. And a quadratic that does not factor over the integers, such as x² + 2x − 1, still has zeros; completing the square gives (x + 1)² − 2, so the zeros are x = −1 ± √2, both irrational. Not factoring "nicely" does not mean no zeros.

About the numbers. Every price, fee, amount and business in this lesson's examples, tasks and questions is made up for practice; none describes a real company, product or offer.

Worked factoring to find zeros: Find the zeros of the function defined by f(x) = x² − 5x − 14. Two numbers with product −14 and sum −5 are −7 and 2, so f(x) = (x − 7)(x + 2). By the zero-product property, f(x) = 0 when x = 7 or x = −2. Check: f(7) = 49 − 35 − 14 = 0 and f(−2) = 4 + 10 − 14 = 0. The factored form reveals both zeros; the standard form conceals them.

Worked factoring with a leading coefficient: Find the zeros of g(x) = 2x² + 7x − 4. Search for (2x + p)(x + q) with pq = −4 and p + 2q = 7. With p = −1 and q = 4, p + 2q = 7 and pq = −4, so g(x) = (2x − 1)(x + 4). Zeros: 2x − 1 = 0 gives x = 1/2; x + 4 = 0 gives x = −4. Check: g(1/2) = 2(1/4) + 7/2 − 4 = 1/2 + 7/2 − 4 = 0, and g(−4) = 32 − 28 − 4 = 0.

Worked completing the square for a minimum: Find the minimum of q(x) = 3x² + 12x + 7. Factor 3 from the x-terms: 3(x² + 4x) + 7. Complete the square inside: x² + 4x = (x + 2)² − 4. Substitute and distribute: 3((x + 2)² − 4) + 7 = 3(x + 2)² − 12 + 7 = 3(x + 2)² − 5. Since 3(x + 2)² ≥ 0 with equality at x = −2, the minimum value is −5 at x = −2. Check: q(−2) = 12 − 24 + 7 = −5, and q(−1) = 3 − 12 + 7 = −2 > −5.

Worked maximum in context: A ball's height in feet is h(t) = −16t² + 48t + 6, an idealized model with no air resistance that applies from launch until the ball lands. Maximum: −16(t² − 3t) + 6 = −16((t − 1.5)² − 2.25) + 6 = −16(t − 1.5)² + 36 + 6 = −16(t − 1.5)² + 42. The maximum height is 42 feet at t = 1.5 seconds. Check: h(1.5) = −36 + 72 + 6 = 42. The vertex form also says how far below the peak the ball is at any time during the flight: 16(t − 1.5)² feet, so at t = 2 it is 16(0.25) = 4 feet below, at 38 feet, and h(2) = −64 + 96 + 6 = 38 confirms it.

Three-forms record: For each of x² − 5x − 14, 2x² + 7x − 4, −2x² + 12x − 5 and 3x² + 12x + 7, make a table with columns standard form, factored form (or "does not factor over the integers" with the zeros from the completed square), vertex form, zeros, extreme value and type (maximum or minimum), and the value at x = 0. Show the expansion check for every factored form and the substitution check for every vertex form. Sketch the parabola for x² − 5x − 14, marking the zeros at −2 and 7, the vertex at (2.5, −20.25), and the y-intercept at −14.

Independent performance: Write a quadratic model for a situation of your own, for example the height of a thrown object, the revenue from a price, or the area of a rectangle with a fixed perimeter. Pose one question that requires the zeros and one that requires the maximum or minimum. Produce the equivalent form that reveals each property, explain what the form shows in the context, and verify each result by substituting into the original expression. Success criteria: the factored form is confirmed by expansion, the completed square is confirmed by expanding it back to the original expression and then spot-checked at the vertex x-value and one neighboring value, the sign of the leading coefficient is used to state whether the extreme is a maximum or a minimum, and each answer is a sentence about the situation rather than only a number.