The reverse process of multiplying polynomials is defined as factoring polynomials. Consider that when we factor a number, we are searching for prime factors that multiply together to give the number. When a polynomial is factorized, usually only the polynomials are broken down to have integer coefficients and constants. Simplest way for factoring is that a common factor for every term. So we can factor out the common factor in the polynomials.
For example, 8=4*2, or 16=4*4
Problems:
Factoring Polynomials using Algebra Tiles:
Example for Factoring Polynomials using Algebra Tiles 1:
Factorize the polynomial x3 – 5x2 – 12x + 36
Solution for Factoring Polynomials using Algebra Tiles 1:
Sum of the coefficients of terms: 1–5–12 + 36 = 18 ≠ 0. ∴ (x–1) is not a factor.
Sum of the coefficients of even degree terms = –5 + 36 = 31
Sum of the coefficients of odd degree terms = 1 – 12 = –11
Since they are not equal we guess that (x + 1) is also not a factor. Let us check whether x – 2
is a factor. By synthetic division method
2 | 1 -5 -12 +36
|
| +2 -6 -36
________________________
1 -3 -18 | 0
_________________________
Since the remainder is 0, (x – 2) is a factor. To find other factors
x2 – 3x – 18 = x2 – 6x + 3x – 18
= x (x–6) + 3 (x–6) = (x + 3) (x – 6)
Therefore, x3 – 5x2 – 12x + 36 = (x–2) (x–6) (x+3)
Sample problem
Example for Factoring Polynomials using Algebra Tiles 2:
Factorize 2x3 + x2 – 5x + 2
Solution for Factoring Polynomials using Algebra Tiles 2:
Since the sum of the coefficients of all the terms: 2 + 1 – 5 + 2 = 5 – 5 = 0
We guess that (x – 1) is a factor.
By synthetic division,
1 | 2 +2 -5 +2
|
| 2 +3 -2
________________________
2 3 -2 | 0
_________________________
Remainder is 0. Quotient is 2x2 + 3x – 2
To find other factors, factorize the quotient,
2x2 + 3x – 2 = 2x2 + 4x – x – 2
= 2x (x + 2) – 1 (x + 2) = (x + 2) (2x – 1)
∴ 2x3 + x2 – 5x + 2 = (x – 1) (x + 2) (2x – 1)
For example, 8=4*2, or 16=4*4
Problems:
Factoring Polynomials using Algebra Tiles:
Example for Factoring Polynomials using Algebra Tiles 1:
Factorize the polynomial x3 – 5x2 – 12x + 36
Solution for Factoring Polynomials using Algebra Tiles 1:
Sum of the coefficients of terms: 1–5–12 + 36 = 18 ≠ 0. ∴ (x–1) is not a factor.
Sum of the coefficients of even degree terms = –5 + 36 = 31
Sum of the coefficients of odd degree terms = 1 – 12 = –11
Since they are not equal we guess that (x + 1) is also not a factor. Let us check whether x – 2
is a factor. By synthetic division method
2 | 1 -5 -12 +36
|
| +2 -6 -36
________________________
1 -3 -18 | 0
_________________________
Since the remainder is 0, (x – 2) is a factor. To find other factors
x2 – 3x – 18 = x2 – 6x + 3x – 18
= x (x–6) + 3 (x–6) = (x + 3) (x – 6)
Therefore, x3 – 5x2 – 12x + 36 = (x–2) (x–6) (x+3)
Sample problem
Example for Factoring Polynomials using Algebra Tiles 2:
Factorize 2x3 + x2 – 5x + 2
Solution for Factoring Polynomials using Algebra Tiles 2:
Since the sum of the coefficients of all the terms: 2 + 1 – 5 + 2 = 5 – 5 = 0
We guess that (x – 1) is a factor.
By synthetic division,
1 | 2 +2 -5 +2
|
| 2 +3 -2
________________________
2 3 -2 | 0
_________________________
Remainder is 0. Quotient is 2x2 + 3x – 2
To find other factors, factorize the quotient,
2x2 + 3x – 2 = 2x2 + 4x – x – 2
= 2x (x + 2) – 1 (x + 2) = (x + 2) (2x – 1)
∴ 2x3 + x2 – 5x + 2 = (x – 1) (x + 2) (2x – 1)