Find the zeros of a factorable polynomial and use them to sketch a rough graph

Learning goal: Identify the zeros of a polynomial when factoring is reasonable, and use the zeros, together with the sign of the polynomial between them and its behavior for large positive and negative x, to construct a rough graph of the function the polynomial defines.

Zeros and factors. A zero of a polynomial p(x) is a number a with p(a) = 0. From the Remainder Theorem, a is a zero exactly when x − a is a factor. So if a polynomial is written as a product of linear factors, its zeros can be read off directly: p(x) = (x + 2)(x − 1)(x − 3) has zeros −2, 1 and 3, because a product is zero exactly when one of its factors is zero, and x + 2 = 0 at x = −2, x − 1 = 0 at x = 1, x − 3 = 0 at x = 3. On the graph of y = p(x), a zero is a point where the curve meets the x-axis, since y = 0 there. This is why zeros are also called x-intercepts.

When factoring is reasonable. The standard is careful: identify zeros when factoring is reasonable. Not every polynomial factors into pieces you can find by hand; x³ − x − 1 has one real zero, but it is not a rational number and cannot be found by the methods of this lesson. The polynomials here factor by the techniques you already know: taking out a common factor, factoring a quadratic, recognizing a difference of squares, or using the Remainder Theorem to test small integer values and divide. If none of those works after a genuine attempt, the honest conclusion is that a rough graph would need technology or a different method, not that the polynomial has no zeros.

From zeros to a rough graph: three ingredients. A rough graph does not need exact heights everywhere. It needs the zeros, the sign of p(x) on each interval between the zeros, and the end behavior, meaning what happens to p(x) as x becomes very large or very negative. The sign on an interval decides whether the curve is above or below the axis there; because a polynomial is continuous and can only change sign at a zero, the sign is constant on each interval, and one test point per interval settles it. The end behavior is decided by the leading term. For a polynomial of odd degree with positive leading coefficient, p(x) is large and positive for large positive x and large and negative for large negative x, so the graph falls on the left and rises on the right. A negative leading coefficient flips both ends. For even degree with positive leading coefficient, both ends rise; with negative leading coefficient, both ends fall.

A fully worked model. Let p(x) = x³ − 2x² − 5x + 6. Test p(1) = 1 − 2 − 5 + 6 = 0, so x − 1 is a factor. Dividing gives x² − x − 6 = (x − 3)(x + 2), and the check is (x − 1)(x − 3)(x + 2) = (x² − 4x + 3)(x + 2) = x³ + 2x² − 4x² − 8x + 3x + 6 = x³ − 2x² − 5x + 6. The zeros are −2, 1 and 3, in order on the number line. They split the axis into four intervals. Test points: at x = −3, p = (−1)(−4)(−6) = −24, negative; at x = 0, p = (2)(−1)(−3) = 6, positive; at x = 2, p = (4)(1)(−1) = −4, negative; at x = 4, p = (6)(3)(1) = 18, positive. The degree is 3 and the leading coefficient is 1, so the graph falls to the left and rises to the right, which agrees with the signs at x = −3 and x = 4. The rough graph starts low on the left, rises through (−2, 0), reaches a hill somewhere between −2 and 1 (the value 8 at x = −1 and 6 at x = 0 show it is up there), comes down through (1, 0), dips to a valley between 1 and 3 (the value −4 at x = 2), and rises through (3, 0) and onward. The diagram plots exactly these values. The precise top of the hill and bottom of the valley are not needed for a rough graph; what is needed is that they exist and lie between the correct zeros.

Repeated factors. In q(x) = −(x + 1)(x − 2)², the factor x − 2 appears twice. The zero 2 is still a zero, but the graph touches the axis at x = 2 without crossing it, because (x − 2)² is never negative and so the sign of q(x) is the same on both sides of 2. Check: q(1) = −(2)(1) = −2 and q(3) = −(4)(1) = −4, both negative. At x = −1 the factor x + 1 appears once, so the sign changes: q(−2) = −(−1)(16) = 16, positive, versus q(0) = −(1)(4) = −4, negative. The rule is that the graph crosses at a zero coming from an odd number of identical factors and touches at a zero coming from an even number. Here the leading term is −x³, so the graph rises to the left and falls to the right: positive at −2, crossing at −1, negative from −1 onward except for the touch at 2.

The misconception to remove. Students sometimes draw the graph as a straight line through the zeros, or draw a hill between every pair of zeros without checking the sign. Between two consecutive zeros where a polynomial crosses the axis, the graph is entirely on one side, and which side is decided by the test point, not by habit. A second error is to assume every polynomial has as many zeros as its degree; x² + 4 has degree 2 and no real zeros, and x⁴ − 16 = (x − 2)(x + 2)(x² + 4) has degree 4 but only two real zeros, so its graph crosses the axis only at −2 and 2.

Reading the graph back. A rough graph built this way answers questions the algebra alone does not make obvious: where p(x) is positive, where it is negative, how many times the curve crosses the axis, and between which zeros it must turn. Those answers come from the zeros and the signs, which is why the standard ties the zeros to the graph. Zeros and signs only constrain the turning points, though: between two neighboring zeros the curve must turn at least once, but zeros and signs alone cannot say exactly where, or whether there are extra turns. For example, x(x² + 1) and x((x − 2)² + 0.1) have the same degree, the same leading coefficient, the single real zero 0 and the same signs on each side of it, yet the first rises everywhere while the second has two turning points. A rough graph is a sketch consistent with the zeros and signs; locating turning points needs more information, such as a table of values or technology.

Worked cubic with a common factor: Sketch f(x) = x³ − 9x. Factor out x: f(x) = x(x² − 9) = x(x − 3)(x + 3). Zeros: −3, 0, 3. Sign test: f(−4) = (−4)(−7)(−1) = −28, negative; f(−1) = (−1)(−4)(2) = 8, positive; f(1) = (1)(−2)(4) = −8, negative; f(4) = (4)(1)(7) = 28, positive. Odd degree, positive leading coefficient: falls left, rises right. The graph rises through (−3, 0), peaks between −3 and 0, falls through the origin, dips between 0 and 3 and rises through (3, 0). The graph is symmetric about the origin, since f(−x) = −f(x).

Worked quartic: Sketch g(x) = x⁴ − 5x² + 4. Treat it as a quadratic in x²: (x² − 1)(x² − 4) = (x − 1)(x + 1)(x − 2)(x + 2). Zeros: −2, −1, 1, 2. Even degree with positive leading coefficient, so both ends rise. Signs: g(−3) = 81 − 45 + 4 = 40, positive; g(−1.5) = 5.0625 − 11.25 + 4 = −2.1875, negative; g(0) = 4, positive; g(1.5) = −2.1875, negative; g(3) = 40, positive. The graph comes down from the upper left, crosses at −2, dips below, crosses up at −1, rises to (0, 4), crosses down at 1, dips, and crosses up at 2 to rise away. Four crossings, three turning points, and a y-intercept of 4.

Worked cubic with a repeated zero: Sketch h(x) = −(x + 1)(x − 2)². Zeros: −1 (single factor, crossing) and 2 (double factor, touching). Expanded, h(x) = −x³ + 3x² − 4, so the leading term is −x³: rises on the left, falls on the right. Signs: h(−2) = −(−1)(16) = 16, positive; h(0) = −(1)(4) = −4, negative; h(3) = −(4)(1) = −4, negative. The graph comes down from the upper left, crosses at (−1, 0), goes below the axis, rises back to touch the axis at (2, 0), and turns down again. The y-intercept is h(0) = −4.

Worked case where factoring stops early: Sketch k(x) = x⁴ − 16 = (x − 2)(x + 2)(x² + 4). Only −2 and 2 are real zeros, because x² + 4 is at least 4 for every real x. Both ends rise (degree 4, positive leading coefficient). k(0) = −16 is negative, so the graph is below the axis between the two zeros and above it outside them, with one valley whose bottom is at (0, −16).

Representation task: For p(x) = x³ − 2x² − 5x + 6, make a sign chart: draw a number line, mark the zeros −2, 1 and 3, choose one test point inside each of the four intervals, compute p at each test point showing the three factor values that multiply to it, and write + or − above each interval. Below the chart, state the end behavior from the leading term x³ and check it agrees with the signs of the outer intervals. Then sketch the rough graph on labeled axes, marking the three zeros, the y-intercept (0, 6) and the approximate location of the two turning points.

Independent performance: Choose a polynomial of degree 3 or 4 that factors by a method from this lesson and has at least one negative zero; write it in expanded form first, then factor it, showing the Remainder Theorem test or the factoring pattern you used. Produce a sign chart and a rough graph. Then write three sentences answering: on which intervals is the polynomial positive; how many times does the graph cross the axis and why is that number at most the degree; and what would change in the sketch if the leading coefficient were made negative. Success criteria: the factored form multiplies back to the expanded form; every interval has a computed test value; the end behavior is justified from the leading term; the sketch crosses the axis at zeros of odd multiplicity (such as a single zero, or the triple zero of (x + 1)³) and touches without crossing at zeros of even multiplicity; and the y-intercept is marked with its value.