Quadratics and Polynomials

Updated 7 October 2026

A quadratic is any equation where the highest power of the unknown is two, and the reason it gets a chapter of its own is that it is the last family of equations that always yields to a formula you can memorize. Everything above it — cubics, quartics, quintics — either factors by luck or gets solved numerically.

There are three ways to solve one, and they are not interchangeable. Each answers a slightly different question, and knowing which to reach for is most of the skill.

How do you solve a quadratic by factoring?

To solve a quadratic by factoring, get everything on one side so the other side is zero, factor into two brackets, and set each equal to zero. $x^2+5x+6=0$x² + 5x + 6 = 0 factors to (x + 2)(x + 3) = 0, so x = −2 or x = −3. The brackets come from two numbers that multiply to the constant and add to the x coefficient: 2 and 3, since 2 × 3 = 6 and 2 + 3 = 5. It works because if two numbers multiply to zero, one of them is zero. With $x^2+x=12$x² + x = 12, writing x(x + 1) = 12 proves nothing; subtract 12 first, then factor $x^2+x-12$x² + x − 12 as (x + 4)(x − 3), so x = −4 or x = 3. Check each root in its original equation: for x = 3, 9 + 3 = 12, and for x = −2, 4 − 10 + 6 = 0.

The lesson on solving a quadratic by factoring drills the search for those two numbers until it is automatic.

What do you do when it will not factor?

When a quadratic will not factor, use the quadratic formula: $x=\left(-b\pm \sqrt{b^2-4ac}\right)$x = (−b ± √(b² − 4ac)) / 2a, with a, b and c taken from $a{x}^2+bx+c=0$ax² + bx + c = 0. For $x^2+4x+1=0$x² + 4x + 1 = 0, a = 1, b = 4 and c = 1. The part under the root is 16 − 4 = 12, so $x=\left(-4\pm \sqrt{12}\right)$x = (−4 ± √12) / 2, which simplifies to $-2\pm \sqrt{3}$−2 ± √3. The formula is completing the square done once, on the general equation. See the quadratic formula, which explains the $\pm$± and the 2a. If b is negative, −b is positive: for $x^2-6x+7=0$x² − 6x + 7 = 0, $b^2-4ac$b² − 4ac is 36 − 28 = 8, so $x=\left(6\pm \sqrt{8}\right)$x = (6 ± √8) / $2=3\pm \sqrt{2}$2 = 3 ± √2. A negative number under the root means no real solution. As a check, the roots add to $-\frac{b}{a}$−b/a: $-2+\sqrt{3}$−2 + √3 and $-2-\sqrt{3}$−2 − √3 add to −4.

$2x^2+3x-4=0\to x=\left(-3\pm \sqrt{9+32}\right)$2x² + 3x − 4 = 0 → x = (−3 ± √(9 + 32)) / $4=\left(-3\pm \sqrt{41}\right)$4 = (−3 ± √41) / 4.

Most real quadratics do not factor. Whole-number factorizations are a small and lucky subset, which is why textbook exercises are full of them and physics problems almost never are.

Where does the quadratic formula come from?

The quadratic formula comes from completing the square, done once on $a{x}^2+bx+c=0$ax² + bx + c = 0 instead of on every problem. Completing the square turns the $x^2$x² and x terms into one squared bracket. For $x^2+8x+3=0$x² + 8x + 3 = 0, halve the 8, since 8 ÷ 2 = 4, and note that ${\left(x+4\right)}^2$(x + 4)² brings in an extra $4^2=16$4² = 16. The equation becomes ${\left(x+4\right)}^2-16+3=0$(x + 4)² − 16 + 3 = 0, which is ${\left(x+4\right)}^2=13$(x + 4)² = 13, so $x=-4\pm \sqrt{13}$x = −4 ± √13. Run the same steps with letters and they produce the formula. See deriving the quadratic formula. When the $x^2$x² term has a coefficient other than 1, divide the whole equation by it first; that division is where the 2a comes from. To check the halving, expand: ${\left(x+4\right)}^2-13$(x + 4)² − 13 is $x^2+8x+16-13$x² + 8x + 16 − 13, and 16 − 13 = 3 is the original constant.

Completing the square rewrites $x^2+6x-1$x² + 6x − 1 as ${\left(x+3\right)}^2-10$(x + 3)² − 10: halve the 6 to get the 3 inside the bracket, then subtract the 9 the bracket brought in with it, along with the original −1. Completing the square is the lesson that makes the halving step feel inevitable rather than magical.

The rewritten form is worth more than the roots. ${\left(x+3\right)}^2-10$(x + 3)² − 10 says at a glance that the minimum value is −10, reached at x = −3, because a squared bracket is never negative and is zero exactly once.

How many solutions does a quadratic have, and can you tell before solving?

A quadratic has two real solutions, one or none, and the discriminant, $b^2-4ac$b² − 4ac, tells you which before you solve. Positive means two, zero means one repeated root, and negative means none. For $2x^2+3x-5=0$2x² + 3x − 5 = 0, the discriminant is 9 + 40 = 49, so there are two roots. For $x^2+2x+5=0$x² + 2x + 5 = 0, it is 4 − 20 = −16, so there is no real solution. The discriminant sits under the square root in the formula: a negative number has no real root, and zero makes the $\pm$± change nothing. See the discriminant and the nature of the roots. Asked for the k that makes the roots equal, set the discriminant to zero: for $x^2+kx+9=0$x² + kx + 9 = 0, $k^2-36=0$k² − 36 = 0, so k = 6 or k = −6. Check k = 6: $x^2+6x+9$x² + 6x + 9 is ${\left(x+3\right)}^2$(x + 3)², and 3 × 3 = 9.

The same quantity answers a question that looks unrelated: whether a quadratic expression is ever negative. If a is positive and the discriminant is negative, the parabola sits entirely above the axis, so the expression is always positive for every real x — a fact used constantly in proofs and in checking that a model cannot produce nonsense.

What do the two roots add up to?

The two roots of a quadratic add to $-\frac{b}{a}$−b/a and multiply to $\frac{c}{a}$c/a, so both can be read off the equation without solving it. For $2x^2-7x+3=0$2x² − 7x + 3 = 0, the sum is $\frac{7}{2}$7/2 and the product is $\frac{3}{2}$3/2. Solving confirms it: the roots are 3 and $\frac{1}{2}$1/2, and $3+\frac{1}{2}=\frac{7}{2}$3 + 1/2 = 7/2 while $3\times \frac{1}{2}=\frac{3}{2}$3 × 1/2 = 3/2. See The sum and product of the roots. Both facts come from the factored form. Expanding $a\left(x-\alpha \right)\left(x-\beta \right)$a(x − α)(x − β) gives $a{x}^2-a\left(\alpha +\beta \right)x+a\alpha \beta$ax² − a(α + β)x + aαβ, and matching it against $a{x}^2+bx+c$ax² + bx + c term by term forces the two relationships. The signs catch people: the sum carries a minus sign and the product does not, so for $x^2+5x+6=0$x² + 5x + 6 = 0 the roots add to −5 and multiply to 6, which makes both negative. Here −2 and −3 fit both facts, since −2 + −3 = −5 and −2 × −3 = 6.

This turns out to be a construction tool. If you know what you want the roots to be, you can build the equation backwards, which is what forming quadratic equations is about — asked for an equation with roots 2 and −5, you write $x^2-\left(\text{sum}\right)x+\left(\text{product}\right)=x^2+3x-10$x² − (sum)x + (product) = x² + 3x − 10 without touching a formula.

It scales up too. A cubic has three roots, and they satisfy three matching relationships; roots and coefficients of a cubic extends the pattern, and transforming the roots of a polynomial shows the trick that follows from it: to find the equation whose roots are double the old ones, you never need the old roots at all.

How do you solve a quadratic inequality?

A quadratic inequality is solved by finding the roots first, then deciding which side of them you want. For $x^2-x-6<0$x² − x − 6 < 0, the expression factors as (x + 2)(x − 3), so the roots are −2 and 3. The parabola $y=x^2-x-6$y = x² − x − 6 opens upward and dips below the axis only between its roots, so the answer is −2 < x < 3. The sign can only change where the curve crosses the axis. See Solving a quadratic inequality. The answer splits in two when the inequality asks for the outside: $x^2-x-6>0$x² − x − 6 > 0 holds for x < −2 or x > 3. A negative $x^2$x² coefficient turns the parabola over, so multiply through by −1 and reverse the sign first. Test a value from each region: x = 0 gives 0 − 0 − 6, a negative, and x = 4 gives 16 − 4 − 6 = 6, a positive.

The same caution applies to rational inequalities, where the denominator can change sign; multiplying up without knowing that sign is the classic way to produce a confident wrong answer.

When does a line meet, touch or miss a curve?

To decide whether a line meets, touches or misses a curve, substitute the line into the curve's equation and read the discriminant of the quadratic. Two solutions mean two crossings, one means a tangent, and none means a miss. For $y=x^2$y = x² and y = 2x + 3, setting $x^2=2x+3$x² = 2x + 3 gives $x^2-2x-3=0$x² − 2x − 3 = 0, with discriminant 4 + 12 = 16, so the line cuts the curve where x = 3 and x = −1. Each solution is the x-coordinate of a shared point. See when a line meets, touches or misses a curve. If the line holds an unknown, such as y = 2x + k, and must be a tangent, set the discriminant to zero and solve for k: 4 + 4k = 0 gives k = −1. Check a meeting point on both graphs: at x = 3, $3^2=9$3² = 9 and 2 × 3 + 3 = 9.

Getting there needs the substitution step itself, which is the subject of a linear equation paired with a quadratic: rearrange the linear equation for one variable, put it into the quadratic, and solve the single-variable equation that results.

Equations that turn into quadratics

A great many equations that look nothing like a quadratic become one after a single rearrangement. Clearing a denominator turns a fractional equation into a quadratic, and squaring both sides does the same for a equation with a square root.

What is the binomial theorem?

The binomial theorem expands a bracket raised to a whole-number power without multiplying it out. ${\left(1+x\right)}^5=1+5x+10x^2+10x^3+5x^4+x^5$(1 + x)⁵ = 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵, and its coefficients are row 5 of Pascal's triangle. Each term takes either the 1 or the x from each of the five brackets. The $x^2$x² term counts the ways to choose two brackets to supply an x, and there are 5 × 4 ÷ 2 = 10 of them. See The binomial theorem. When the first term is not 1, its powers enter every coefficient. In ${\left(2+x\right)}^3$(2 + x)³ the coefficients 1, 3, 3, 1 are multiplied by falling powers of 2, giving $8+12x+6x^2+x^3$8 + 12x + 6x² + x³, because 3 × 4 = 12 and 3 × 2 = 6. Putting x = 1 checks the row: 1 + 5 + 10 + 10 + 5 + 1 = 32, five 2s multiplied together.

The theorem then extends past whole-number powers, which is where it stops being an algebra convenience and becomes an approximation tool. The binomial series for a negative or fractional index turns things like $\sqrt{1+x}$√(1 + x) and $\frac{1}{\left(1+x\right)}$1/(1 + x) into infinite series that are accurate for small x, and expanding (a + bx) to a rational power handles the case where the bracket does not conveniently start with a 1 — you factor the a out first, and the condition for validity comes along with it.

How do you factor a cubic?

To factor a cubic, find one root by trial, divide out the factor it gives, and factor what is left. For $x^3-6x^2+11x-6$x³ − 6x² + 11x − 6, try x = 1: 1 − 6 + 11 − 6 = 0, so (x − 1) is a factor. Dividing leaves $x^2-5x+6$x² − 5x + 6, which is (x − 2)(x − 3), so the cubic is (x − 1)(x − 2)(x − 3). The factor theorem justifies the trial: if x = a makes a polynomial zero, then (x − a) is a factor. Only divisors of the constant term need trying. See The factor and remainder theorem. When the substitution does not give zero, the value is the remainder instead: at x = 4, 64 − 96 + 44 − 6 = 6, so dividing by (x − 4) leaves 6. Check the factored form at x = 0, where the brackets give −1 × −2 × −3 = −6, the constant term.

With whole-number coefficients, a cubic ending in −6 has only $\pm 1$±1, $\pm 2$±2, $\pm 3$±3, $\pm 6$±6 worth testing.

What if there is no exact answer?

When no exact solution can be written down, rearrange it into the shape x = g(x), start from a rough guess, and feed each output back in as the next input. The values usually settle toward a root. For $x^2=2$x² = 2, one rearrangement is $x=\left(x+\frac{2}{x}\right)/2$x = (x + 2/x) / 2. Starting from 1 gives $\left(1+2\right)÷2=\frac{3}{2}$(1 + 2) ÷ 2 = 3/2, and the next step gives $\left(\frac{3}{2}+\frac{4}{3}\right)÷2=\frac{17}{12}$(3/2 + 4/3) ÷ 2 = 17/12, already close to $\sqrt{2}$√2. A root is a fixed point: put it in and the same number comes out, so nearby values are pulled toward it. See Solving equations by iteration. Not every rearrangement settles. Written as $x=\frac{2}{x}$x = 2/x, the same equation sends 1 to 2 and 2 back to 1 forever, so try another shape when values jump about. Substitute the result into the original equation: 17 × 17 = 289 and 2 × 144 = 288, so $\frac{17}{12}$17/12 squared is just over 2.

This matters more than it first appears. Beyond the fourth degree there is no general formula in radicals, so iteration is not a fallback for the lazy: for most polynomial equations that arise in practice, it is the method.

The mistakes worth naming

• Solving before setting it to zero. (x + 2)(x + 3) = 2 tells you nothing about the brackets. The zero-product property needs the zero.
• Losing the negative root. $\sqrt{9}$√9 is 3, but "$x^2=9$x² = 9" has two answers. The square root symbol and the act of undoing a square are different things.
• Forgetting to subtract what completing the square added. ${\left(x+3\right)}^2$(x + 3)² is not $x^2+6x$x² + 6x; it carries a spare 9.
• Trusting a squared equation. Every radical equation ends with substituting the answers back in.

The natural next step from here is graphs of equations and coordinate geometry, which is the same parabola seen from the other side, and functions and rational functions, where completing the square reappears as a graph transformation. Related guides: Complex Numbers and Matrices and Markov Chains.