MATH FACTORING - USING FACTOR THEOREM AND SYNTHETIC DIVISION, EXAMPLES

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Please study
Method of Factoring Polynomial before Math Factoring
if you have not already done so.

There we discussed about
Remainder Theorem, Factor Theorem
and explained the Method of factoring
with Example.

Here, we will apply the method to solve problems.
The knowledge of Synthetic Division is made use of here.
That knowledge is a prerequisite here.
So, please learn the method before proceeding further.

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Example 1 of Math Factoring

Solve the following problem on Math Factoring

Factorize x4 + 5x3 + 5x2 - 5x - 6

Solution to Example 1 of Math Factoring

Let f(x) = x4 + 5x3 + 5x2 - 5x - 6
Here constant term = -6. Its factors are +1,-1,+2, -2, +3, -3, +6, -6. So we check witha = +1,-1,+2, -2, +3, -3, +6, -6 whether f(a) is zero or not.
f(1) = 1 + 5 + 5 - 5 - 6 = 0
x -(1) = (x - 1) is a factor of f(x).
Now let us divide f(x) = (x4 + 5x3 + 5x2 - 5x - 6 ) by (x - 1) usingSynthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)

1  |   +1    +5    +5    -5    -6
   |   +0    +1    +6    +11   +6  
   |------------------------------------
   |   +1    +6    +11   +6    +0 

∴ Quotient = x3 + 6x2 + 11x + 6
Thus, (x4 + 5x3 + 5x2 - 5x - 6) ÷ (x - 1) = x3 + 6x2 + 11x + 6
∴ f(x) = (x - 1)(x3 + 6x2 + 11x + 6)

Let p(x) = x3 + 6x2 + 11x + 6This is a third Degree Polynomial, which is to be factorized using Facor Theorem.
Here constant term = 6. Its factors are +1,-1,+2, -2, +3, -3, +6, -6. So we check witha = +1,-1,+2, -2, +3, -3, +6, -6 whether f(a) is zero or not.
p(1) is obviously not zero, as all the coefficients are positive.
p(-1) = (-1)3 + 6(-1)2 + 11(-1) + 6 = -1 + 6 -11 +6 = 0
x -(-1) = (x + 1) is a factor of p(x).
Now let us divide p(x) = (x3 + 6x2 + 11x + 6) by (x + 1) usingSynthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)

-1  |   +1    +6    +11    +6
    |   +0    -1    -5     -6  
    |-----------------------------
    |   +1    +5    +6     +0 

∴ Quotient = (+1)x2 + (+5)x + (6) = x2 + 5x + 6
Thus, p(x) ÷ (x + 1) = x2 + 5x + 6
∴ p(x) = (x + 1)(x2 + 5x + 6)
We can factorize x2 + 5x + 6 using the knowledge of factoring Polynomials of second degree.
x2 + 5x + 6 = x2 + 2x + 3x + 6= x(x + 2) + 3(x + 2) = (x + 2)(x + 3)
∴ p(x) = (x + 1)(x + 2)(x + 3)
∴ f(x) = (x - 1)p(x) = (x - 1)(x + 1)(x + 2)(x + 3).
Thus, x4 + 5x3 + 5x2 - 5x - 6
= (x - 1)(x + 1)(x + 2)(x + 3).
Thus Factoring Polynomial x4 + 5x3 + 5x2 - 5x - 6 by using Factor Theorem gave the Factors as(x - 1)(x + 1)(x + 2)(x + 3). Ans.
Thus, the problem on Math Factoring is solved.

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Example 2 of Math Factoring

Solve the following problem on Math Factoring

Factorize 4x4 - 12x3 + 7x2 + 3x - 2

Solution to Example 2 of Math Factoring :

Let f(x) = 4x4 - 12x3 + 7x2 + 3x - 2
Here constant term = -2. Its factors are +1,-1,+2, -2. So we check witha = +1,-1,+2, -2, whether f(a) is zero or not.
f(1) = 4 - 12 + 7 + 3 - 2 = 0 < BR> ⇒ x - (1) = (x - 1) is a factor of f(x).
Now let us divide f(x) = (4x4 - 12x3 + 7x2 + 3x - 2) by (x - 1) usingSynthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)

1  |   +4    -12    +7    +3    -2
   |   +0    +4     -8    -1    +2  
   |------------------------------------
   |   +4    -8     -1    +2    +0 

∴ Quotient = 4x3 - 8x2 - x + 2
Thus, (4x4 - 12x3 + 7x2 + 3x - 2) ÷ (x - 1) = 4x3 - 8x2 - x + 2
∴ f(x) = (x - 1)(4x3 - 8x2 - x + 2)

Let p(x) = 4x3 - 8x2 - x + 2This is a third Degree Polynomial, which is to be factorized using Facor Theorem.
Here constant term = 2. Its factors are +1,-1,+2, -2. So we check witha = +1,-1,+2, -2, whether f(a) is zero or not.
p(1) = 4 - 8 -1 + 2 = -3 ≠ 0
p(-1) = 4(-1)3 - 8(-1)2 -(-1) + 2 = -4 - 8 + 1 + 2 ≠ 0
p(2) = 4(2)3 - 8(2)2 -(2) + 2 = 32 - 32 - 2 + 2 = 0.
x -(2) = (x - 2) is a factor of p(x).
Now let us divide p(x) = (4x3 - 8x2 - x + 2) by (x - 2) usingSynthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)

2  |   +4    -8    -1    +2
   |   +0    +8    +0    -2  
   |-----------------------------
   |   +4    +0    -1    +0 

∴ Quotient = (+4)x2 + (0)x - (1) = 4x2 - 1
Thus, p(x) ÷ (x - 2) = 4x2 - 1
∴ p(x) = (x - 2)(4x2 - 1)
But 4x2 - 1 = (2x)2 - (1)2 is differenceof two squares which as per Formula is the Product of Sum and diffrence.
∴ 4x2 - 1 = (2x)2 - (1)2 = (2x + 1)(2x - 1)
∴ p(x) = (x - 2)(2x + 1)(2x - 1)
∴ f(x) = (x - 1)p(x) = (x - 1)(x - 2)(2x + 1)(2x - 1).
Thus, 4x4 - 12x3 + 7x2 + 3x - 2
= (x - 1)(x - 2)(2x + 1)(2x - 1).
Thus Factoring Polynomial 4x4 - 12x3 + 7x2 + 3x - 2 by using Factor Theorem gave the Factors as(x - 1)(x - 2)(2x + 1)(2x - 1). Ans.

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Exercise on Math Factoring

Solve the Following Problems on Math Factoring :

  1. Do the Math Factoring of x4 + 2x3 - 7x2 - 8x + 12
  2. Do the Math Factoring of x4 + 3x3 - 7x2 - 27x - 18

For Answers of these problems on Math Factoring,
See at the bottom of the Page.

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Answers to Exercise on Math Factoring

Answers to the problems in Exercise
on Math Factoring are given below.

  1. (x - 1)(x - 2)(x + 2)(x + 3)
  2. (x - 3)(x + 1)(x + 2)(x + 3)