Solve x³ − 19x + 30 = 0
Polynomial equation, worked out line by line the way a teacher would write it.
Answer
| Solutions | x = −5, x = 2, x = 3 |
Step-by-step solution
11 steps-
1 Givenx^{3} - 19 x + 30 = 0
Solve for x.
-
2 Rational root theorem: any rational root is ±(divisor of 30)/(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,\; -5,\; 5,\; -6,\; 6,\; -10,\; 10,\; -15,\; 15,\; -30,\; 30 \} -
3 Test the candidates: x = 2 makes the polynomial 0, so (x - 2) is a factor
If substituting a number makes the polynomial 0, the factor theorem says that (x - 2) is a factor. Dividing it out leaves a polynomial one degree lower, which is easier to solve.
30 - 19 \cdot 2 + 2^{3} = 0 -
4 Divide the polynomial by (x - 2) (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 - 2\right) \left(x^{2} + 2 x - 15\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 = 2 and c = -15, signs included, shows which method fits: factoring, taking a square root or the quadratic formula.
a = 1,\quad b = 2,\quad c = -15 -
6 Factor the trinomial: find two numbers whose product is c = -15 and whose sum is b = 2
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 = -15 and add up to b = 2.
List the pairs of numbers whose product is -15, with their signs, and pick the pair whose sum is 2.
(-3) \cdot 5 = -15,\qquad (-3) + 5 = 2 -
7 Write the factored form\left(x - 3\right) \left(x + 5\right) = 0
-
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 - 3 = 0\quad \text{or} \quad x + 5 = 0 -
9 Add 3 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 = 3Move the constant terms to the right.
-
10 Subtract 5 from both sidesx = -5
Move the constant terms to the right.
-
11 Check\begin{aligned}x = -5:\quad 0 = 0\quad\checkmark\\ x = 2:\quad 0 = 0\quad\checkmark\\ x = 3:\quad 0 = 0\quad\checkmark\end{aligned}
Substituting each solution back makes both sides equal.
Check by substitution
| x | Left side | Right side | |
|---|---|---|---|
| −5 | 0 | 0 | ✓ |
| 2 | 0 | 0 | ✓ |
| 3 | 0 | 0 | ✓ |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.