Solve x³ − 13x − 12 = 0
Polynomial equation, worked out line by line the way a teacher would write it.
Answer
| Solutions | x = −3, x = −1, x = 4 |
Step-by-step solution
11 steps-
1 Givenx^{3} - 13 x - 12 = 0
Solve for x.
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2 Rational root theorem: any rational root is ±(divisor of 12)/(divisor of 1)
If a polynomial with whole-number coefficients has a root p/q in lowest terms, then p divides the constant term and q divides the leading coefficient. That gives a short list of candidates to test, instead of guessing.
x \in \{ -1,\; 1,\; -2,\; 2,\; -3,\; 3,\; -4,\; 4,\; -6,\; 6,\; -12,\; 12 \} -
3 Test the candidates: x = -1 makes the polynomial 0, so (x + 1) is a factor
If substituting a number makes the polynomial 0, the factor theorem says that (x + 1) is a factor. Dividing it out leaves a polynomial one degree lower, which is easier to solve.
\left(-1\right)^{3} - 12 - 13 \left(-1\right) = 0 -
4 Divide the polynomial by (x + 1) (synthetic division)
Dividing by the factor just found splits the polynomial into that factor times a polynomial one degree lower. Synthetic division is a shortcut for this long division that only works with the coefficients.
\left(x + 1\right) \left(x^{2} - x - 12\right) = 0 -
5 This is a quadratic in standard form ax² + bx + c = 0
Every quadratic equation can be arranged as ax² + bx + c = 0. Reading off a = 1, b = -1 and c = -12, signs included, shows which method fits: factoring, taking a square root or the quadratic formula.
a = 1,\quad b = -1,\quad c = -12 -
6 Factor the trinomial: find two numbers whose product is c = -12 and whose sum is b = -1
The goal is to write x² + bx + c as (x + p)(x + q). Multiplying that out gives x² + (p + q)x + p·q, so p and q must multiply to c = -12 and add up to b = -1.
List the pairs of numbers whose product is -12, with their signs, and pick the pair whose sum is -1.
(-4) \cdot 3 = -12,\qquad (-4) + 3 = -1 -
7 Write the factored form\left(x - 4\right) \left(x + 3\right) = 0
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8 Zero product property: a product is 0 only when one of its factors is 0
0 is the only number with this property: if a·b = 0, then a = 0 or b = 0. That is why the equation was first rearranged to … = 0 and factored: now each factor can be set to 0 on its own, giving a simpler equation for each.
x - 4 = 0\quad \text{or} \quad x + 3 = 0 -
9 Add 4 to both sides
An equation stays true as long as you do the same thing to both sides, like adding the same weight to both pans of a balance. Adding or subtracting the same amount on both sides moves a term to the other side, where it appears with the opposite sign.
x = 4Move the constant terms to the right.
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10 Subtract 3 from both sidesx = -3
Move the constant terms to the right.
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11 Check\begin{aligned}x = -3:\quad 0 = 0\quad\checkmark\\ x = -1:\quad 0 = 0\quad\checkmark\\ x = 4:\quad 0 = 0\quad\checkmark\end{aligned}
Substituting each solution back makes both sides equal.
Check by substitution
| x | Left side | Right side | |
|---|---|---|---|
| −3 | 0 | 0 | ✓ |
| −1 | 0 | 0 | ✓ |
| 4 | 0 | 0 | ✓ |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.