Step-by-step explanation:
Example 1
Solve the equation x3 − 3x2 – 2x + 4 = 0
We put the numbers that are factors of 4 into the equation to see if any of them are correct.
f(1) = 13 − 3×12 – 2×1 + 4 = 0 1 is a solution
f(−1) = (−1)3 − 3×(−1)2 – 2×(−1) + 4 = 2
f(2) = 23 − 3×22 – 2×2 + 4 = −4
f(−2) = (−2)3 − 3×(−2)2 – 2×(−2) + 4 = −12
f(4) = 43 − 3×42 – 2×4 + 4 = 12
f(−4) = (−4)3 − 3×(−4)2 – 2×(−4) + 4 = −100
The only integer solution is x = 1. When we have found one solution we don’t really need to test any other numbers because we can now solve the equation by dividing by (x − 1) and trying to solve the quadratic we get from the division.
Now we can factorise our expression as follows:
x3 − 3x2 – 2x + 4 = (x − 1)(x2 − 2x − 4) = 0
It now remains for us to solve the quadratic equation.
x2 − 2x − 4 = 0
We use the formula for quadratics with a = 1, b = −2 and c = −4.
We have now found all three solutions of the equation x3 − 3x2 – 2x + 4 = 0. They are: eftirfarandi:
x = 1
x = 1 + Ö5
x = 1 − Ö5
Example 2
We can easily use the same method to solve a fourth degree equation or equations of a still higher degree. Solve the equation f(x) = x4 − x3 − 5x2 + 3x + 2 = 0.
First we find the integer factors of the constant term, 2. The integer factors of 2 are ±1 and ±2.
f(1) = 14 − 13 − 5×12 + 3×1 + 2 = 0 1 is a solution
f(−1) = (−1)4 − (−1)3 − 5×(−1)2 + 3×(−1) + 2 = −4
f(2) = 24 − 23 − 5×22 + 3×2 + 2 = −4
f(−2) = (−2)4 − (−2)3 − 5×(−2)2 + 3×(−2) + 2 = 0 we have found a second solution.
The two solutions we have found 1 and −2 mean that we can divide by x − 1 and x + 2 and there will be no remainder. We’ll do this in two steps.
First divide by x + 2
Now divide the resulting cubic factor by x − 1.
We have now factorised
f(x) = x4 − x3 − 5x2 + 3x + 2 into
f(x) = (x + 2)(x − 1)(x2 − 2x − 1) and it only remains to solve the quadratic equation
x2 − 2x − 1 = 0. We use the formula with a = 1, b = −2 and c = −1.
Now we have found a total of four solutions. They are:
x = 1
x = −2
x = 1 +
x = 1 −
Sometimes we can solve a third degree equation by bracketing the terms two by two and finding a factor that they have in common.
