Solve x³ + 5x² − 4x − 20 = 0
Polynomial equation, worked out line by line the way a teacher would write it.
Answer
| Solutions | x = −5, x = −2, x = 2 |
Step-by-step solution
11 steps-
1 Givenx^{3} + 5 x^{2} - 4 x - 20 = 0
Solve for x.
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2 Rational root theorem: any rational root is ±(divisor of 20)/(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,\; -4,\; 4,\; -5,\; 5,\; -10,\; 10,\; -20,\; 20 \} -
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.
5 \left(-2\right)^{2} - 20 - 4 \left(-2\right) + \left(-2\right)^{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} + 3 x - 10\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 = 3 and c = -10, signs included, shows which method fits: factoring, taking a square root or the quadratic formula.
a = 1,\quad b = 3,\quad c = -10 -
6 Factor the trinomial: find two numbers whose product is c = -10 and whose sum is b = 3
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 = -10 and add up to b = 3.
List the pairs of numbers whose product is -10, with their signs, and pick the pair whose sum is 3.
(-2) \cdot 5 = -10,\qquad (-2) + 5 = 3 -
7 Write the factored form\left(x - 2\right) \left(x + 5\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 - 2 = 0\quad \text{or} \quad x + 5 = 0 -
9 Add 2 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 = 2Move the constant terms to the right.
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10 Subtract 5 from both sidesx = -5
Move the constant terms to the right.
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11 Check\begin{aligned}x = -5:\quad 0 = 0\quad\checkmark\\ x = -2:\quad 0 = 0\quad\checkmark\\ x = 2:\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 | ✓ |
| 2 | 0 | 0 | ✓ |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.