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23 December,2014

In mathematical terms, all cubic equations have either one root or three real roots. The general cubic equation is,

**ax ^{3}+ bx^{2}+ cx+d= 0**

The coefficients of a, b, c and d are real or complex numbers with a not equals to zero (aÂ â‰ Â 0). It must have the term **x ^{3}** Â in it, or else it will not be a cubic equation. But any or all of b, c and d can be zero.

The examples of cubic equations are,

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **No 1,Â Â Â ****x ^{3}**

**Â No 2,Â Â Â 4x ^{3}+ 57=0Â **

**Â No 3,Â Â x ^{3}+ 9x=0Â Â **

Unlike quadratic equation which may have no real solution; a cubic equation always has at least one real root. The prior strategy of solving a cubic equation is to reduce it to a quadratic equation, and then solve the quadratic by usual means, either by factorizing or using a formula.

Always try to find the solution of cubic equations with the help of the general equation,

**ax ^{3}+ bx^{2}+ cx+d= 0**

A cubic equation should, therefore, must be re-arranged into its standard form,

For example,

**x ^{2}+ 4x-1 =Â 6/x**

You can see the equation is not written in standard form, you need to multiply the â€˜xâ€™ to eliminate the fraction and get cubic equation, after doing so, you will end up with

**x ^{3}+ 4x^{2}– x = 6**

Then you subtract 6 from both sides in order to get â€˜0â€™ on the right side, so you will come up,

**x ^{3}+ 4x^{2}– x- 6 = 0**

What is factor theorem? If you divide a polynomial p(x) by a factor x â€“ a of that polynomial, then, you will end up with zero as the remainder,

**p(x) = (x â€“ a)q(x) + r(x)**

If **x â€“ a** is indeed a factor of p(x), then the remainder after division by **x â€“ a** will be zero.

**p(x) = (x â€“ a)q (x)**

Here is a problem,

**x ^{3}– 5x^{2}– 2x+24 = 0**

WithÂ **x= – 2Â **a solution.

Factor theorem says that ifÂ **Â x = – 2**Â is a solution of this equation, thenÂ **x+2**Â is a factor of this whole expression. So,

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **Â x ^{3}– 5x^{2}– 2x+24 = 0**

can be written in the form,

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **Â Â (x+2)( x ^{2}+ ax+b)=0Â Â **

WhereÂ *a*Â andÂ *b*Â are numbers,

You need to find out what are the values ofÂ *a*Â andÂ *b*Â here by using synthetic division.

First, you need to look at the coefficients of the original cubic equation, which areÂ **1, -5, -2**Â andÂ **24.**

To the right of the vertical you need to write down the known root,**Â x = -2**

Now multiply number (1) that just brought down, by the known rootÂ **-2**, as a result isÂ **-2**, you mention the result in the other line, like

The numbers in the second column are added, so giving us,

Then recently written number**Â 7**Â is multiplied by the known root,Â **– 2**,

As 14 comes as a result, you need to write it down on the second row over the line,

Like previously the numbers in this column added,Â **(14 â€“ 2 = 12)**

And you need to go on with the process,

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **Â Â **

When you have zero at the bottom row, it gives the confirmation that **x = – 2** is a root of the original cubic. At this stage, you got the first three numbers in the bottom row as the coefficients in the quadratic,

x^{2}– 7x+12

Hence, you reduced your cubic to,

(x+2)(x^{2}Â – 7x + 12) =0

**Step 8**

After applying the quadratic term, the equation comes like this,

(x +2) (x â€“ 3) (x â€“ 4) = 0

Resulting, you get the solution asÂ **x = -2**Â orÂ **3Â **orÂ **4**.

The equation is,

x^{3}– 7x-6=0

(-1)^{3 – }7(-1) -6

The result comes as zero, so it is proved that x +1 is a factor and the cubic can be written in the form,

(x+1)(x^{2}+ax+b)= 0

After applying the synthetic division, like above example, you will take the coefficients of the original cubic equation, which areÂ **1, 0, -7**Â andÂ **-6**, you need to write down the know root x = -1 to the right of the vertical line, giving us,

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **Â **

Multiply the brought down numberÂ **1**Â by the known rootÂ **x = -1**, and put down the resultÂ **(-1)**Â at the second row, like this,

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

The numbers of the second column are added to the first column, giving us,

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

And continue the process adding numbers in this column, until you find**Â â€˜0â€™**Â as a result,

As you add more numbers to the second column by following the synthetic division process, you will come with,

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

When you come up with zero, it is confirmation thatÂ **x = -1**,

Hence, you get the coefficients in the quadratic as the first three numbers in the bottom row, so the quadratic is,

**x ^{2Â }-x – 6**

Hence, the cubic reduced to quadratic,

**(x+1)(x ^{2}-x- 6) =0**

The factorized result is,

**(x +1)(x â€“ 3)(x + 2) = 0**

You can get three solutions to the cubic equation areÂ **x = -2, -1Â **or**Â 3**

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