Solving a Cubic Completely

Find one root by trial

Solve $x^3-2x^2-5x+6=0$x³ − 2x² − 5x + 6 = 0. Write $f\left(x\right)=x^3-2x^2-5x+6$f(x) = x³ − 2x² − 5x + 6. There is no formula to reach for, so the first job is to find one root by trial.

The factor theorem narrows the search. If the roots are whole numbers p, q and s, then f(x) = (x − p)(x − q)(x − s), whose constant term is −pqs. So pqs = −6, and each whole-number root divides 6. The candidates are $\pm 1$±1, $\pm 2$±2, $\pm 3$±3 and $\pm 6$±6.

Try them in turn. f(1) = 1 − 2 − 5 + 6 = 0, so x = 1 is a root and x − 1 is a factor. Had the first try failed, the search would carry on: f(2) = 8 − 8 − 10 + 6 = −4 and f(−1) = −1 − 2 + 5 + 6 = 8, so neither 2 nor −1 is a root.

Divide by the factor

Dividing the cubic by x − 1 leaves a quadratic. By long division: $x^3/x=x^2$x³ / x = x², and $x^2\left(x-1\right)=x^3-x^2$x²(x − 1) = x³ − x²; subtracting from $x^3-2x^2$x³ − 2x² leaves $-x^2$−x², and bringing down −5x gives $-x^2-5x$−x² − 5x. Then $-x^2/x=-x$−x² / x = −x, and $-x\left(x-1\right)=-x^2+x$−x(x − 1) = −x² + x; subtracting leaves −6x, and bringing down 6 gives −6x + 6. Then $-6x/x=-6$−6x / x = −6, and −6(x − 1) = −6x + 6, which leaves 0.

So $x^3-2x^2-5x+6=\left(x-1\right)\left(x^2-x-6\right)$x³ − 2x² − 5x + 6 = (x − 1)(x² − x − 6). The remainder is 0, as the factor theorem said it would be.

Or match the coefficients

The same quadratic comes from matching coefficients. Write $x^3-2x^2-5x+6=\left(x-1\right)\left(x^2+bx+c\right)$x³ − 2x² − 5x + 6 = (x − 1)(x² + bx + c) and multiply out the right side: $x^3+\left(b-1\right)x^2+\left(c-b\right)x-c$x³ + (b − 1)x² + (c − b)x − c.

The $x^2$x² terms give b − 1 = −2, so b = −1. The constants give −c = 6, so c = −6. The x terms are then a check: c − b = −6 + 1 = −5, which matches. The quadratic is $x^2-x-6$x² − x − 6.

Finish on the quadratic

Factor the quadratic: $x^2-x-6=\left(x-3\right)\left(x+2\right)$x² − x − 6 = (x − 3)(x + 2), because −3 × 2 = −6 and −3 + 2 = −1. So $x^3-2x^2-5x+6=\left(x-1\right)\left(x-3\right)\left(x+2\right)$x³ − 2x² − 5x + 6 = (x − 1)(x − 3)(x + 2).

The cubic is 0 when any bracket is 0, so the roots are x = 1, x = 3 and x = −2. Check the two new roots by substitution: f(3) = 27 − 18 − 15 + 6 = 0 and f(−2) = −8 − 8 + 10 + 6 = 0. All three divide 6, as the search predicted.

The curve $y=x^3-2x^2-5x+6$y = x³ − 2x² − 5x + 6 crosses the x-axis three times, at the dots: x = −2, x = 1 and x = 3. Those are the three roots, one found by trial and two from the quadratic.

The same method on two more cubics

Solve $x^3+2x^2-5x-6=0$x³ + 2x² − 5x − 6 = 0. The candidates divide 6. f(1) = 1 + 2 − 5 − 6 = −8, not 0. f(2) = 8 + 8 − 10 − 6 = 0, so x − 2 is a factor. Dividing leaves $x^2+4x+3=\left(x+1\right)\left(x+3\right)$x² + 4x + 3 = (x + 1)(x + 3), so the roots are 2, −1 and −3.

Solve $x^3-3x^2-6x+8=0$x³ − 3x² − 6x + 8 = 0. The candidates divide 8. f(1) = 1 − 3 − 6 + 8 = 0, so x − 1 is a factor. Dividing leaves $x^2-2x-8=\left(x-4\right)\left(x+2\right)$x² − 2x − 8 = (x − 4)(x + 2), so the roots are 1, 4 and −2. Check: f(4) = 64 − 48 − 24 + 8 = 0.

When the quadratic does not factor

Solve $x^3-3x^2-3x+1=0$x³ − 3x² − 3x + 1 = 0. The candidates are $\pm 1$±1. f(1) = 1 − 3 − 3 + 1 = −4, but f(−1) = −1 − 3 + 3 + 1 = 0, so x + 1 is a factor. Dividing leaves $x^2-4x+1$x² − 4x + 1, which has no whole-number factors.

Use the quadratic formula: $x=\left(4\pm \sqrt{16-4}\right)$x = (4 ± √(16 − 4)) / $2=\left(4\pm 2\sqrt{3}\right)$2 = (4 ± 2√3) / $2=2\pm \sqrt{3}$2 = 2 ± √3, because $\sqrt{12}=\sqrt{4}\times \sqrt{3}=2\sqrt{3}$√12 = √4 × √3 = 2√3. So the roots are −1, $2+\sqrt{3}\approx 3.732$2 + √3 ≈ 3.732 and $2-\sqrt{3}\approx 0.268$2 − √3 ≈ 0.268. Only one root was found by trial; the other two are irrational, and the quadratic formula finds them.

Check $2+\sqrt{3}$2 + √3 in the quadratic: ${\left(2+\sqrt{3}\right)}^2-4\left(2+\sqrt{3}\right)+1=7+4\sqrt{3}-8-4\sqrt{3}+1=0$(2 + √3)² − 4(2 + √3) + 1 = 7 + 4√3 − 8 − 4√3 + 1 = 0.

The curve $y=x^3-3x^2-3x+1$y = x³ − 3x² − 3x + 1. It crosses the x-axis at x = −1, the root found by trial, and between the gridlines at the two roots from the quadratic formula: $2-\sqrt{3}$2 − √3, about 0.268, and $2+\sqrt{3}$2 + √3, about 3.732.

When there is only one real root

Solve $x^3-1=0$x³ − 1 = 0. f(1) = 0, so x − 1 is a factor, and dividing leaves $x^2+x+1$x² + x + 1. Its discriminant is $1^2-4\times 1\times 1=-3$1² − 4 × 1 × 1 = −3, which is negative, so the quadratic has no real roots. The cubic has one real root, x = 1.

A cubic always has at least one real root, because its graph runs from far below the x-axis to far above it. It has three when the quadratic left over has two.

The usual mistakes

Calling a candidate a root without testing it. For $x^3-2x^2-5x+6$x³ − 2x² − 5x + 6, 2 divides 6, but f(2) = −4, so x − 2 is not a factor.

Getting the sign of the factor wrong. The root x = 1 gives the factor x − 1, not x + 1; f(−1) = 8, so x + 1 is not a factor.

Changing the signs of the roots from the quadratic. $x^2-x-6=\left(x-3\right)\left(x+2\right)$x² − x − 6 = (x − 3)(x + 2) has roots 3 and −2, not −3 and 2.

Stopping at two roots, or giving the root already found as one of the other two. The quadratic holds the other two roots, so the cubic $x^3-2x^2-5x+6$x³ − 2x² − 5x + 6 has three: 1, 3 and −2.

A frame from 28 m of rod

In the application below, the cubic is $2x^3-7x^2+9=0$2x³ − 7x² + 9 = 0. Its leading coefficient is 2, so the candidates include halves as well as whole numbers. The first root found is x = −1, which cannot be a length, and the two lengths come from the quadratic left after dividing it out.