FACTORING POLYNOMIAL USING FACTOR THEOREM AND SYNTHETIC DIVISION, EXAMPLES

Method of Factoring Polynomial
if you have not already done so.

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 Factoring Polynomial

Solve the following problem on Factoring Polynomial

Factorize x3 + 5x2 + 7x + 3

Solution to Example 1 of Factoring Polynomial

Let f(x) = x3 + 5x2 + 7x + 3
Here constant term = 3. Its factors are +1, -1, +3, -3. So we check witha = +1, -1, +3, -3, whether f(a) is zero or not.
f(1) is obviously not zero, as all the coefficients are positive.
f(-1) = (-1)3 + 5(-1)2 + 7(-1) + 3 = -1 + 5(1) -7 + 3 = 0
x -(-1) = (x + 1) is a factor of f(x).
After one factor is arrived at, we can divide the given Polynomialwith this factor, using the method explained in
Dividing Polynomials.

However, we have a more convenient method calledSYNTHETIC DIVISION which is explained below.
To divide x3 + 5x2 + 7x + 3 by (x + 1)
Using Synthetic Division:

```-1  |   +1    +5    +7    +3
|   +0    -1    -4    -3
|-----------------------------
|   +1    +4    +3    +0
```

Step1: Equate the Divisor (x + 1) to zero and find x (= -1) and write thisvalue of x on the left of the column.
Step2: Arrange the Dividend in the descending powers of the variable (x)and write the coefficients of the Dividend in descending order to the right side of the column as first row. If any term is missing, write '0' as its coefficient.
Step3: write '0' as the first element of the second row (below the firstelement of the first row)
Step4: Add the two numbers of the first column(+1 + 0) and write its value (+1) as the first number in the third row.
Step5: Multiply the first number in the third row (+1) with the number on the left of the column (-1) and write the result (+1 x -1 = -1) as the second element of the second row.
Step6: Add the two numbers in the 2nd column [ +5 + (-1) ] and write the result ( = +4 ) as the second element of the third row.
Step7: Multiply the second number in the third row (+4) with the number on the left of the column (-1) and write the result (+4 x -1 = -4) as the third element of the second row.
Step8: Add the two numbers in the 3rd column [ +7 + (-4) ] and write the result ( = +3 ) as the third element of the third row.
Step9: Multiply the third number in the third row (+3) with the number on the left of the column (-1) and write the result (+3 x -1 = -3) as the fourth element of the second row.
Step10: Add the two numbers in the 4th column [ +3 + (-3) ] and write the result ( = +0 ) as the fourth element of the third row.

Note: You might have noted, steps 8,9,10 are repetition of steps 5, 6, 7.

Now, the third row numbers ( +1, +4, +3; last '0' is ignored.) form the coefficients of the Quotient in descending order starting with the power of the variable one less than the highest power of the Dividend.

∴ Here, the Quotient = x2 + 4x + 3.
Thus, (x3 + 5x2 + 7x + 3) ÷ (x + 1) = x2 + 4x + 3
x3 + 5x2 + 7x + 3 = (x + 1)(x2 + 4x + 3)
We can factorize x2 + 4x + 3 using the knowledge of factoring Polynomials of second degree.
x2 + 4x + 3 = x2 + x + 3x + 3= x(x + 1) + 3(x + 1) = (x + 1)(x + 3)
x3 + 5x2 + 7x + 3 = (x + 1)(x + 1)(x + 3)
= (x + 1)2(x + 3).
Thus Factoring Polynomial x3 + 5x2 + 7x + 3by using Factor Theorem gave the Factors as(x + 1)2(x + 3). Ans.

Example 1 of Factoring Polynomial is thus solved.

Example 2 of Factoring Polynomial

Solve the following problem on Factoring Polynomial

Factorize 4x3 + 8x2 - 6x - 12

Solution to Example 2 of Factoring Polynomial

Let f(x) = 4x3 + 8x2 - 6x - 12
Here constant term = -12. Its factors are +1,-1,+2, -2, +3, -3, +4, -4, +6, -6, +12, -12. So we check witha = +1,-1,+2, -2, +3, -3, +4, -4, +6, -6, whether f(a) is zero or not.
f(1) = 4 + 8 -6 -12 = -6 ≠ 0; f(-1) = 4(-1)3 + 8(-1)2 -6(-1) -12 = -4 + 8 +6 -12 = -2 ≠ 0;
f(2) = 4(2)3 + 8(2)2 - 6(2) - 12 ≠ 0;
f(-2) = 4(-2)3 + 8(-2)2 - 6(-2) - 12 = -32 +32 +12 -12 = 0;
x -(-2) = (x + 2) is a factor of f(x).
Now let us divide f(x) = (4x3 + 8x2 - 6x - 12) by (x + 2) usingSynthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)

```-2  |   +4    +8    -6    -12
|   +0    -8    +0    +12
|-----------------------------
|   +4    +0    -6    +0
```

∴ Quotient = (+4)x2 + (0)x + (-6) = 4x2 - 6
Thus, (4x3 + 8x2 - 6x - 12) ÷ (x + 2) = 4x2 - 6
∴ 4x3 + 8x2 - 6x - 12 = (x + 2)(4x2 - 6)
But 4x2 - 6 = (2x)2 - (√6)2 is differenceof two squares which as per Formula is the Product of Sum and diffrence.
∴ 4x2 - 6 = (2x)2 - (√6)2 = (2x + √6)(2x - √6)

∴ (4x3 + 8x2 - 6x - 12) = (x + 2)(2x + √6)(2x - √6).
Thus Factoring Polynomial 4x3 + 8x2 - 6x - 12 by using Factor Theorem gave the Factors as(x + 2)(2x + √6)(2x - √6). Ans.

Example 2 of Factoring Polynomial is thus solved.

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Exercise on Factoring Polynomial

Solve the Following Problems on Factoring polynomials :

1. Factorize x3 - 4x2 + 5x - 2
2. Factorize x3 - x2 - 5x - 3

For Answers See at the bottom of the Page.

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Answers to Exercise on Factoring Polynomial

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

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

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