Solving cubic equations with the help of factor theorem

1

Introduction
Learn to Solve Cubic Equations In mathematical
terms, all cubic equations have either one root
or three real roots. The general cubic equation
is, ax3 bx2 cxd 0 The coefficients of a, b,
c and d are real or complex numbers with a not
equals to zero (a ? 0). It must have the
term x3   in it, or else it will not be a cubic
equation.
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Introduction
The coefficients of a, b, c and d are real or
complex numbers with a not equals to zero
(a ? 0). It must have the term x3   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,                           No
1,    x3 3a3 3 a2  a3 b0  
No 2,    4x3 570   
No 3,    x3 9x0   
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Strategy to Solve Cubic Equation

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, ax3 bx2 cxd 0
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Strategy to Solve Cubic Equation

A cubic equation should, therefore, must be
re-arranged into its standard form, For
example, x2 4x-1   6/x
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Strategy to Solve Cubic Equation

Step 1 You can see the equation is not written
in standard form, you need to multiply the x to
eliminate the fraction and get cubic equation,
after doing so, you will end up with x3 4x2 x
6
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Strategy to Solve Cubic Equation

Step 2 Then you subtract 6 from both sides in
order to get 0 on the right side, so you will
come up, x3 4x2 x- 6 0
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Solving Cubic Equations with the help of Factor
Theorem

What is factor theorem? If you divide a
polynomial p(x) by a factor x a of that
polynomial, then, you will end up with zero as
the remainder, p(x) (x a)q(x) r(x)
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Solving Cubic Equations with the help of Factor
Theorem

If x a is indeed a factor of p(x), then the
remainder after division by x a will be
zero. p(x) (x a)q (x) Here is a
problem, x3 5x2 2x24 0 With x 2 a
solution.
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Solving Cubic Equations with the help of Factor
Theorem

If x a is indeed a factor of p(x), then the
remainder after division by x a will be
zero. p(x) (x a)q (x) Here is a
problem, x3 5x2 2x24 0 With x 2 a
solution.
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Solving Cubic Equations with the help of Factor
Theorem

Factor theorem says that if  x 2 is a
solution of this equation, then x2 is a factor
of this whole expression.                        
                                                  
                                                  
                                       
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Solving Cubic Equations with the help of Factor
Theorem

Step 1 First, you need to look at the
coefficients of the original cubic equation,
which are 1, -5, -2 and 24.
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Solving Cubic Equations with the help of Factor
Theorem

Step 2 Now multiply number (1) that just brought
down, by the known root -2, as a result is -2,
you mention the result in the other line, like
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Solving Cubic Equations with the help of Factor
Theorem

Step 3 The numbers in the second column are
added, so giving us,
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Solving Cubic Equations with the help of Factor
Theorem

Step 4 Then recently written number 7 is
multiplied by the known root,  2, As 14 comes as
a result, you need to write it down on the second
row over the line,
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Solving Cubic Equations with the help of Factor
Theorem

Step 5 Like previously the numbers in this column
added, (14 2 12)
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Solving Cubic Equations with the help of Factor
Theorem

Step 6 And you need to go on with the process,
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Solving Cubic Equations with the help of Factor
Theorem

Step 7 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, x2 7x12 Hence,
you reduced your cubic to,
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Solving Cubic Equations with the help of Factor
Theorem

(x2)(x2  7x 12) 0 Step 8 After applying the
quadratic term, the equation comes like this, (x
2) (x 3) (x 4) 0 Resulting, you get the
solution as x -2 or 3 or 4.
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Solving Cubic Equations with the help of Factor
Theorem

Another Example The equation is, x3
7x-60 Step 1 You can simply try x 1, after
putting the value of x, you will get, (-1)3
 7(-1) -6
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Solving Cubic Equations with the help of Factor
Theorem

Step 2 After applying the synthetic division,
like above example, you will take the
coefficients of the original cubic equation,
which are 1, 0, -7 and -6, you need to write down
the know root x -1 to the right of the vertical
line, giving us,
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Solving Cubic Equations with the help of Factor
Theorem

Step 3 Multiply the brought down number 1 by the
known root x -1, and put down the
result (-1) at the second row, like this,
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Solving Cubic Equations with the help of Factor
Theorem

Step 4 The numbers of the second column are
added to the first column, giving us,
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Solving Cubic Equations with the help of Factor
Theorem

Step 5 As you add more numbers to the second
column by following the synthetic division
process, you will come with,
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Solving Cubic Equations with the help of Factor
Theorem

Step 6 Hence, the cubic reduced to
quadratic, (x1)(x2-x- 6) 0 The factorized
result is, (x 1)(x 3)(x 2) 0 You can get
three solutions to the cubic equation are x -2,
-1 or 3
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