**Answer:**

**Step-by-step explanation:**

How can we solve equations such as the cubic equation shown here?

x3 − x2 – 4x + 4 = 0

There is an extremely complicated formula for solving cubic equations. Some calculators have this formula built in and therefore can be used to solve cubic equations.

We are going to learn how these equations can be solved by factorising. If the equation has solutions that are integers a, b and c then we can factorise the equation as follows:

x3 − x2 – 4x + 4 = (x − a)(x − b)(x − c) = 0

Multiplying the brackets together we see that the constant term, 4, must be the number we get when we multiply a, b and c together.

abc = 4

All the solutions a, b and c must be factors of 4 so there are not many whole numbers that we need to consider.

We have only the following possibilities:

±1, ±2 and ±4

We’ll examine each of these numbers to find which ones are solutions of the equation.

f(1) = 13 − 12 – 4×1 + 4 = 0 1 is a solution

f(−1) = (−1)3 − (−1)2 – 4×(−1) + 4 = 6

f(2) = 23 − 22 – 4×2 + 4 = 0 2 is a solution

f(−2) = (−2)3 − (−2)2 – 4×(−2) + 4 = 0 −2 is a solution

We have now found three solutions so we don’t need to try 4 and −4 as a cubic equation has a maximum of three solutions.

These three numbers give us the values of a, b and c and we can factorise the equation.

x3 − x2 – 4x + 4 = (x − 1)(x − 2)(x + 2) = 0

This method involves finding integers that are factors of ( can be divided into ) the constant term and then testing whether these integers are solutions of the equation.

Unfortunately we cannot assume that the solutions of a third degree equation are all integers.

However, if we can find one integer solution, lets say it is x = a then, by the remainder theorem , we know that (x − a) is a factor of the equation. We can find another factor, a quadratic factor, by division. We can then solve the quadratic equation by using the formula for solving quadratics.

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 −

hopefully this helps!:)