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How to Solve Cubic Equation pdf

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Introduction<br />

Learn <strong>to</strong> <strong>Solve</strong> <strong>Cubic</strong> <strong>Equation</strong>s<br />

In mathematical terms, all cubic equations have<br />

either one root or three real roots. The general<br />

cubic equation is,<br />

ax 3 + bx 2 + cx+d= 0<br />

The coefficients of a, b, c and d are real or<br />

complex numbers with a not equals <strong>to</strong> zero<br />

(a ≠ 0). It must have the term x 3 in it, or else it<br />

will not be a cubic equation.


Introduction<br />

The coefficients of a, b, c and d are real or<br />

complex numbers with a not equals <strong>to</strong> zero<br />

(a ≠ 0). It must have the term x 3 in it, or else it<br />

will not be a cubic equation. But any or all of b, c<br />

and d can be zero.<br />

The examples of cubic equations are,<br />

No 1, x 3 + 3a 3 + 3 a 2 + a 3 – b=0<br />

No 2, 4x 3 + 57=0<br />

No 3, x 3 + 9x=0


Strategy <strong>to</strong> <strong>Solve</strong> <strong>Cubic</strong><br />

<strong>Equation</strong><br />

Unlike quadratic equation which may have no<br />

real solution; a cubic equation always has at<br />

least one real root. The prior strategy of solving<br />

a cubic equation is <strong>to</strong> reduce it <strong>to</strong> a quadratic<br />

equation, and then solve the quadratic by usual<br />

means, either by fac<strong>to</strong>rizing or using a formula.<br />

Always try <strong>to</strong> find the solution of cubic<br />

equations with the help of the general equation,<br />

ax 3 + bx 2 + cx+d= 0


Strategy <strong>to</strong> <strong>Solve</strong> <strong>Cubic</strong><br />

<strong>Equation</strong><br />

A cubic equation should, therefore, must be rearranged<br />

in<strong>to</strong> its standard form,<br />

For example,<br />

x 2 + 4x-1 = 6/x


Strategy <strong>to</strong> <strong>Solve</strong> <strong>Cubic</strong><br />

<strong>Equation</strong><br />

Step 1<br />

You can see the equation is not written in<br />

standard form, you need <strong>to</strong> multiply the ‘x’ <strong>to</strong><br />

eliminate the fraction and get cubic equation,<br />

after doing so, you will end up with<br />

x 3 + 4x 2 – x = 6


Strategy <strong>to</strong> <strong>Solve</strong> <strong>Cubic</strong><br />

<strong>Equation</strong><br />

Step 2<br />

Then you subtract 6 from both sides in order <strong>to</strong><br />

get ‘0’ on the right side, so you will come up,<br />

x 3 + 4x 2 – x- 6 = 0


Solving <strong>Cubic</strong> <strong>Equation</strong>s with<br />

the help of Fac<strong>to</strong>r Theorem<br />

What is fac<strong>to</strong>r theorem? If you divide a<br />

polynomial p(x) by a fac<strong>to</strong>r x – a of that<br />

polynomial, then, you will end up with zero as<br />

the remainder,<br />

p(x) = (x – a)q(x) + r(x)


Solving <strong>Cubic</strong> <strong>Equation</strong>s with<br />

the help of Fac<strong>to</strong>r Theorem<br />

If x – a is indeed a fac<strong>to</strong>r of p(x), then the<br />

remainder after division by x – a will be zero.<br />

Here is a problem,<br />

With x= – 2 a solution.<br />

p(x) = (x – a)q (x)<br />

x 3 – 5x 2 – 2x+24 = 0


Solving <strong>Cubic</strong> <strong>Equation</strong>s with<br />

the help of Fac<strong>to</strong>r Theorem<br />

If x – a is indeed a fac<strong>to</strong>r of p(x), then the<br />

remainder after division by x – a will be zero.<br />

Here is a problem,<br />

With x= – 2 a solution.<br />

p(x) = (x – a)q (x)<br />

x 3 – 5x 2 – 2x+24 = 0


Solving <strong>Cubic</strong> <strong>Equation</strong>s with<br />

the help of Fac<strong>to</strong>r Theorem<br />

Fac<strong>to</strong>r theorem says that if x = – 2 is a solution<br />

of this equation, then x+2 is a fac<strong>to</strong>r of this<br />

whole expression.


Solving <strong>Cubic</strong> <strong>Equation</strong>s with<br />

Step 1<br />

the help of Fac<strong>to</strong>r Theorem<br />

First, you need <strong>to</strong> look at the coefficients of the<br />

original cubic equation, which are 1, -5, -<br />

2 and 24.


Solving <strong>Cubic</strong> <strong>Equation</strong>s with<br />

the help of Fac<strong>to</strong>r Theorem<br />

Step 2<br />

Now multiply number (1) that just brought<br />

down, by the known root -2, as a result is -2, you<br />

mention the result in the other line, like


Solving <strong>Cubic</strong> <strong>Equation</strong>s with<br />

the help of Fac<strong>to</strong>r Theorem<br />

Step 3<br />

The numbers in the second column are added,<br />

so giving us,


Solving <strong>Cubic</strong> <strong>Equation</strong>s with<br />

the help of Fac<strong>to</strong>r Theorem<br />

Step 4<br />

Then recently written number 7 is multiplied by<br />

the known root, – 2,<br />

As 14 comes as a result, you need <strong>to</strong> write it<br />

down on the second row over the line,


Solving <strong>Cubic</strong> <strong>Equation</strong>s with<br />

the help of Fac<strong>to</strong>r Theorem<br />

Step 5<br />

Like previously the numbers in this column<br />

added, (14 – 2 = 12)


Solving <strong>Cubic</strong> <strong>Equation</strong>s with<br />

the help of Fac<strong>to</strong>r Theorem<br />

Step 6<br />

And you need <strong>to</strong> go on with the process,


Solving <strong>Cubic</strong> <strong>Equation</strong>s with<br />

Step 7<br />

the help of Fac<strong>to</strong>r Theorem<br />

When you have zero at the bot<strong>to</strong>m row, it gives<br />

the confirmation that x = – 2 is a root of the<br />

original cubic. At this stage, you got the first<br />

three numbers in the bot<strong>to</strong>m row as the<br />

coefficients in the quadratic,<br />

x 2 – 7x+12<br />

Hence, you reduced your cubic <strong>to</strong>,


Solving <strong>Cubic</strong> <strong>Equation</strong>s with<br />

the help of Fac<strong>to</strong>r Theorem<br />

(x+2)(x 2 – 7x + 12) =0<br />

Step 8<br />

After applying the quadratic term, the equation<br />

comes like this,<br />

(x +2) (x – 3) (x – 4) = 0<br />

Resulting, you get the solution as x = -2 or 3 or 4.


Solving <strong>Cubic</strong> <strong>Equation</strong>s with<br />

the help of Fac<strong>to</strong>r Theorem<br />

Another Example:<br />

The equation is,<br />

x 3 – 7x-6=0<br />

Step 1 You can simply try x = – 1, after putting<br />

the value of x, you will get,<br />

(-1) 3 – 7(-1) -6


Solving <strong>Cubic</strong> <strong>Equation</strong>s with<br />

Step 2<br />

the help of Fac<strong>to</strong>r Theorem<br />

After applying the synthetic division, like above<br />

example, you will take the coefficients of the<br />

original cubic equation, which are 1, 0, -7 and -6,<br />

you need <strong>to</strong> write down the know root x = -1 <strong>to</strong><br />

the right of the vertical line, giving us,


Solving <strong>Cubic</strong> <strong>Equation</strong>s with<br />

Step 3<br />

the help of Fac<strong>to</strong>r Theorem<br />

Multiply the brought down number 1 by the<br />

known root x = -1, and put down the result (-<br />

1) at the second row, like this,


Solving <strong>Cubic</strong> <strong>Equation</strong>s with<br />

Step 4<br />

the help of Fac<strong>to</strong>r Theorem<br />

The numbers of the second column are added <strong>to</strong><br />

the first column, giving us,


Solving <strong>Cubic</strong> <strong>Equation</strong>s with<br />

Step 5<br />

the help of Fac<strong>to</strong>r Theorem<br />

As you add more numbers <strong>to</strong> the second column<br />

by following the synthetic division process, you<br />

will come with,


Solving <strong>Cubic</strong> <strong>Equation</strong>s with<br />

Step 6<br />

the help of Fac<strong>to</strong>r Theorem<br />

Hence, the cubic reduced <strong>to</strong> quadratic,<br />

(x+1)(x 2 -x- 6) =0<br />

The fac<strong>to</strong>rized result is,<br />

(x +1)(x – 3)(x + 2) = 0<br />

You can get three solutions <strong>to</strong> the cubic<br />

equation are x = -2, -1 or 3


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