SAT Polynomial Equations on Desmos: Find Every Root

On Desmos, the real solutions of a polynomial equation are the x-intercepts of its graph, so you can count and read them without expanding. Graph y = the polynomial, click the curve, and read the gray points on the x-axis. The same graph shows where the polynomial is positive and which roots repeat.

Watch: SAT Math Tricks: Sum of Coefficients in One Step (Short)

SAT Math Tricks: Sum of Coefficients in One Step

You can follow along on the College Board version of the calculator on our Desmos on the SAT practice page. Bluebook, the app you take the SAT in, has Desmos built in, and College Board allows a calculator on the whole Math section. Below, you watch two SAT-style polynomial questions solved step by step, finish two more with help, and then solve four on your own.

The rule: four things the graph tells you

• Solutions. A real solution of p(x) = 0 is an x-value where the graph of y = p(x) meets the x-axis. Count the x-intercepts to count the distinct real solutions.
• Sign. p(x) > 0 wherever the graph sits above the x-axis, and p(x) < 0 wherever it sits below.
• Multiplicity. If the graph crosses the axis at a root, that factor appears an odd number of times. If the graph touches the axis and turns back, the factor appears an even number of times. The graph of y = (x - 2)²(x + 1) touches at x = 2 and crosses at x = -1.
• Sum of roots. For ax² + bx + c = 0 and for ax³ + bx² + cx + d = 0, the roots add to -b/a. That sum counts a repeated root once for each time it repeats, and it includes roots that are not real. If a question asks for the sum of the distinct real solutions, list them from the graph instead.

Desmos's help pages describe the two moves you need. Click anywhere on a curve, and gray dots appear at intercepts, intersections, maximums and minimums; click a dot to show its coordinates. Type a letter where a number would go, and you can add a slider for it.

Example 1, worked for you: count the solutions

How many distinct real solutions does x⁴ - 5x² + 4 = 0 have?

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SAT-style question written for this guide.

Line 1: y = x^4 - 5x^2 + 4

I type the left side as y = and leave the 0 out. The solutions of the equation are the x-values where this graph reaches y = 0, and y = 0 is the x-axis.

Before I click, I predict. A degree 4 polynomial has at most 4 real roots, so the answer is 0, 1, 2, 3 or 4.

I click the curve. Gray dots sit on the x-axis at x = -2, -1, 1 and 2, and the graph crosses the axis at each one. Four crossings on a degree 4 polynomial use up all 4 roots, so none of them repeats.

Then I check by hand. Treat x² as one unknown: x⁴ - 5x² + 4 = (x² - 1)(x² - 4) = (x + 1)(x - 1)(x + 2)(x - 2). Four different factors give four solutions, so the answer is 4.

Example 2, worked for you: where is the polynomial positive?

f(x) = 3(x - a)(x - b)(x - c), where a, b and c are constants and a < b < c. The graph of y = f(x) in the xy-plane passes through the point (r, 8). Which of the following could be true?

I. r < a
II. b < r < c
III. r > c

A) I only
B) III only
C) I and II only
D) II and III only

SAT-style question written for this guide.

The question uses letters, and Desmos needs numbers. So I pick values in the required order: a = 3, b = 7 and c = 13. Any three values in that order give the same sign pattern.

Line 1: f(x) = 3(x - 3)(x - 7)(x - 13)
Line 2: y = 8

The point (r, 8) is on the graph, so f(r) = 8. Line 2 draws every point with a y-value of 8. Wherever that horizontal line meets the curve, the x-value is a possible r.

I click the curve and look at where it meets y = 8. It meets the line three times: twice between 3 and 7, and once to the right of 13. It never meets the line to the left of 3 or between 7 and 13, because the curve is below the x-axis there.

So r sits between a and b, or to the right of c. Statement I puts r where f is negative, so it can't be true. Statement II does the same. Statement III can be true. The answer is B, III only.

By hand: a cubic with a positive leading coefficient and three distinct roots is negative to the left of its smallest root, then changes sign at each root. The pattern runs negative, positive, negative, positive. f(r) = 8 is positive, so r is between a and b or greater than c.

Example 3, you finish it: sum of solutions with a constant

(3x + p)(5x² - 45)(2x² - 16x + 6p) = 0

In the given equation, p is a positive constant. The sum of all the solutions to the equation is 20/3. What is the value of p?

SAT-style question written for this guide.

Step 1: go factor by factor. A product equals 0 when one of its factors equals 0.

• 3x + p = 0 gives x = -p/3.
• 5x² - 45 = 5(x - 3)(x + 3), which gives x = 3 and x = -3. Those two add to 0.
• 2x² - 16x + 6p = 0 has two roots that add to -b/a = -(-16)/2 = 8. You don't need the roots themselves.

Step 2 is yours. Add the three pieces, set the total equal to 20/3, and solve for p.

Check your answer: -p/3 + 0 + 8 = 20/3, so -p/3 = 20/3 - 24/3 = -4/3, and p = 4. Confirm it on Desmos. Type y = (3x + 4)(5x^2 - 45)(2x^2 - 16x + 24) and click the curve. The x-intercepts are -3, -4/3, 2, 3 and 6 (if a label reads -1.333, that is -4/3). The five values add to 20/3, and no two are equal, so the formula and the graph agree.

Example 4, you finish it: use a slider

x³ - 3x + k = 0

In the given equation, k is a positive constant. The equation has exactly two distinct real solutions. What is the value of k?

SAT-style question written for this guide.

Step 1: make k a slider.

Line 1: y = x^3 - 3x + k

Add the slider for k. Desmos's help pages say you can click the numbers at either end of a slider to change its range, so set it to run from 0 to 5.

Step 2: watch the count. Drag k up from 0. For small k the curve crosses the x-axis three times. For large k it crosses once. The count can only be two at the moment in between, when a turning point of the curve sits on the x-axis.

Step 3 is yours. Click the curve to show the gray dot at its low point on the right and read the x-value. Then find the k that makes y = 0 at that x.

Check your answer: the low point is at x = 1, where y = 1 - 3 + k = k - 2. It sits on the axis when k - 2 = 0, so k = 2. By hand, x³ - 3x + 2 = (x - 1)²(x + 2). The graph touches the axis at x = 1 and crosses at x = -2: exactly two distinct solutions. The high point, at x = -1, has y = k + 2, which reaches the axis only when k = -2, and the question says k is positive.

This example also shows why -b/a needs care. The x² coefficient is 0, so the roots add to 0 when you count the repeated root twice: 1 + 1 + (-2) = 0. The two distinct solutions add to 1 + (-2) = -1.

Your turn: four problems

Solve all four before you read the answers.

1. (x + 2)(x² - 9)(x² - 4x + 3) = 0. List the distinct real solutions. How many are there?
2. f(x) = -2(x - 1)(x - 4)(x - 7). For what values of x is f(x) > 0?
3. How many real solutions does x⁴ + 1 = 0 have?
4. In the equation (2x - k)(x² + 6x + 9) = 0, k is a constant. The sum of the distinct solutions is 2. What is the value of k?

SAT-style practice problems written for this guide.

Answers

1. Four: -3, -2, 1 and 3. x² - 9 = (x - 3)(x + 3) and x² - 4x + 3 = (x - 1)(x - 3), so x = 3 comes from two factors. On Desmos the graph touches the axis at x = 3 and crosses it at the other three.
2. x < 1 or 4 < x < 7. The leading coefficient, -2, is negative, so the graph sits above the axis on the far left and changes sign at 1, 4 and 7: positive, negative, positive, negative.
3. Zero. x⁴ is never negative, so x⁴ + 1 is at least 1. The graph of y = x⁴ + 1 stays above the x-axis.
4. k = 10. x² + 6x + 9 = (x + 3)², so -3 is one solution, counted once. The other solution is x = k/2. Then k/2 - 3 = 2, so k = 10. Check: the solutions are 5 and -3, and 5 + (-3) = 2.

When algebra is faster

College Board's calculator tips say some Math questions are better done without a calculator. With polynomials, a sum question like Example 3 is the clearest case: -b/a on one factor gives you 8 in a single step. A factored equation also shows its solutions on sight. (x - 1)(x - 4)(x - 7) = 0 has solutions 1, 4 and 7, and you can graph it afterward as a check.

Mistakes to avoid

• Counting a repeated root twice. A graph that touches the axis and turns back marks one solution. Count it once for "distinct solutions", and twice only inside -b/a.
• Deciding the count from the default view. A root can sit off screen. Zoom out before you answer, and compare the count with the degree: a degree 4 polynomial can't have 5 roots.
• Answering with a decimal. If a root reads -1.333, the exact value is -4/3. Match the fraction to the answer choices, or substitute it back to confirm.

Practice next

Work through Advanced Math: Solution Counts, Matching Terms and Absolute Value Equations, free lesson notes on this topic. For mixed Math practice, open the TTA SAT question bank. Related Desmos guides: the sum of solutions formula, quadratic functions on Desmos, and absolute value functions on Desmos.

Checked on September 26, 2026 against College Board's SAT Suite calculator policy and the Desmos Help Center pages FAQs and Sliders and Movable Points in a Graph. We recomputed every answer on this page with SymPy and NumPy. This version corrects the slider example and two practice answers from an earlier version of this guide. The Test Advantage is independent of College Board and Desmos.