Factoring Polynomials

Sometimes we can transform the polynomials into product of several multipliers - polynomials or monomials. Such identical transformation is called factoring polynomials.

In this case we say, that polynomial is divided into each of these factors.

Let′s consider some ways of factoring polynomials.

  1. Factoring out the common factor.

    These transformations are result of distributive law (we need only write these law "from right to left"): `ac+bc=c(a+b)`.

    Example 1. Factoring the following: `28x^3-35x^4`.

    `28x^3-35x^4=7x^3*4-7x^3*5x=7x^3(4-5x)`.

    In general when we factor out variables, we need to factor out every variables with smallest exponent that polynomial has. If all coefficients of the polynomial are integer numbers then we need to factor out absolute value of greatest common divisor of all coefficients of the polynomial.

  2. Using of formulas of reduced multiplication.

    If we read formulas of reduced multiplication from "right to left", in many cases they are very useful for factoring polynomials.

    Example 2. Factor the following:`x^6-1`.

    `x^6-1=(x^3)^2-1^2` . We use the formula of the difference of the squares and obtain `(x^3+1)(x^3-1)`. Then we use the formulas of sum of cubes and difference of cubes and find `(x+1)(x^2-x+1)(x-1)(x^2+x+1)`. So, `x^6-1=(x+1)(x-1)(x^2-x+1)(x^2+x+1)` .

    Example 3. Factor the following: `4a^4b^3+16a^3b^4+16a^2b^5`.

    At the beginning we factor out the common factor. For this we find the greatest common divisor of coefficients 4, 16, 16 and the least exponents `a` and `b`. We will obtain `4a^2b^3(a^2+4ab+4b^2)`. But by the formula 2 we have `a^2+4ab+4b^2=(a+2b)^2` , therefore we finally obtain `4a^4b^3+16a^3b^4+16a^2b^5=4a^2b^3(a+2b)^2` .

  3. The grouping method.

    It is based on the fact, that commutative and associative laws of addition allow to group the members of polynomial by the various ways.

    Sometimes such grouping can be so, that after factoring out the common factors in each group in the brackets remain the same polynomial, which by turn as the common multiplier can be factoring out.

    Example 4. Factor the following: `x^3-3x^2+5x-15`.

    We make the following grouping: `(x^3-3x^2)+(5x-15)`. In the first group we factor out the common multiplier `x^2` , in the second-the common multiplier 5. We obtain `x^2(x-3)+5(x-3)`. Now the polynomial `(x-3)` as the common multiplier we factoring out: `(x-3)(x^2+5)`. Thus, `x^3-3x^2+5x-15=(x-3)(x^2+5)`.

    Example 5. Factor the following: `20x^2+3yz-15xy-4xz`.

    `20x^2+3yz-15xy-4xz=(20x^2-15xy)+(3yz-4xz)=`

    `=5x(4x-3y)-z(4x-3y)=(4x-3y)(5x-z)`.

    Example 6. Factor the following: `a^2-7ab+12b^2` . Here neither groupping will not give common factor. In such cases we sometimes can represent some member of polynomial as some sum and then apply groupping method. In given example we represent `-7ab` in the form of `-3ab-4ab`. Then we obtain `a^2-7ab+12b^2=a^2-3ab-4ab+12b^2=(a^2-3ab)-(4ab-12b^2)=a(a-3b)--4b(a-3b)=(a-3b)(a-4b)`

    Example 7. Factor the following: `x^4+4y^4`. We add and and subtract the monomial `4x^2y^2` : `x^4+4y^4=(x^4+4x^2y^2+4y^4)-4x^2y^2=(x^2+2y^2)^2-(2xy)^2=(x^2+2y^2-2xy)(x^2+2y^2+2xy)` .