Algebraic Expression Factorization

algebraic expressions are expressions that display numbers and variables, and make the algebraic expression factorization means to write the expression as a multiplication of two or more terms.

Factoring algebraic expressions can make many algebraic calculations easier, because when we factor, we can simplify the expression. But how to factor algebraic expressions?

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To factor algebraic expressions, we use the techniques that we will see next.

factoring by evidence

Factoring by evidence consists of highlighting a common term in the algebraic expression.

This common term can be just a number, a variable, or a multiplication of the two, that is, it is a monomial.

Example:

factor the expression \dpi{120} \mathrm{3xy - 2x^2}.

Note that in both terms of this expression the variable appears \dpi{120} \mathrm{x}, so let's put it in evidence:

\dpi{120} \mathrm{3xy - 2x^2 x\cdot (3y-2x)}

Factoring by grouping

At factoring bygrouping, we group the terms that have a factor in common. Then we bring the common factor to the fore.

Thus, the common factor is a polynomial and no longer a monomial, as in the previous case.

Example:

factor the expression \dpi{120} \mathrm{ax^2 - 2ay + 5x^2 - 10y}.

Note that the expression is formed by a sum of several terms and that, in some terms, appears \dpi{120} \mathrm{x^2} and in others it appears \dpi{120} \mathrm{y}.

Let's rewrite the expression, grouping these terms together:

\dpi{120} \mathrm{ax^2 + 5x^2 - 10y - 2ay}

Let's put the variables \dpi{120} \mathrm{x^2} It is \dpi{120} \mathrm{y} in evidence:

\dpi{120} \mathrm{x^2(a+5)-y (2a+10)}

Now, see that the term \dpi{120} \mathrm{y (2y + 10)} can be rewritten as \dpi{120} \mathrm{y (2a + 2\cdot 5)}, from which we can put the number 2 in evidence as well:

\dpi{120} \mathrm{x^2(a+5)-2y (a+5)}

like the polynomial \dpi{120} \mathrm{(a+5)} appears in both terms, we can put it in evidence once more:

\dpi{120} \mathrm{(a+5)\cdot (x^2-2y)}

Therefore, \dpi{120} \mathrm{ax^2 - 2ay + 5x^2 - 10y (a+5)\cdot (x^2 - 2y)}.

Factoring the difference of two squares

If the expression is a difference of two squares, it can be written as the product of the sum of the bases and the difference of the bases. It is one of notable products:

\dpi{120} \mathrm{(a^2 - b^2) (a +b)\cdot (a-b)}

Example:

factor the expression \dpi{120} \mathrm{81 - 4x^2}.

Note that this expression can be rewritten as \dpi{120} \mathrm{9^2 - (2x)^2}, that is, it is a difference of two square terms, whose bases are 9 and 2x.

So let's write the expression as the product of the sum of the bases and the difference of the bases:

\dpi{120} \mathrm{81 - 4x^2 (9+2x)\cdot (9-2x)}

Factoring the perfect square trinomial

In factoring the perfect square trinomial, we also use the notable products and write the expression as the square of the sum or square of the difference between two terms:

\dpi{120} \mathrm{a^2 + 2ab+b^2 (a + b)\cdot (a+b) (a+b)^2}
\dpi{120} \mathrm{a^2 - 2ab+b^2 (a - b)\cdot (a-b) (a-b)^2}

Example:

factor the expression \dpi{120} \mathrm{x^2 + 22y + 121}.

Note that the expression is a perfect square trinomial, as \dpi{120} \mathrm{\sqrt{x^2} x}, \dpi{120} \sqrt{121}11 It is \dpi{120} \mathrm{2\cdot x\cdot 11 22y}.

Then we can factor the expression, writing it as the square of the sum of two terms:

\dpi{120} \mathrm{x^2 + 22y + 121 (x + 11)\cdot (x + 11) (x + 11)^2}

Perfect cube factorization

If the expression is a perfect cube, we factor by writing the expression as the sum cube or difference cube.

\dpi{120} \mathrm{a^3 + 3a^2b + 3b^2a + b^3 (a + b)^3 }
\dpi{120} \mathrm{a^3 - 3a^2b + 3b^2a + b^3 (a - b)^3 }

Example:

factor the expression \dpi{120} \mathrm{x^3 + 6x^2 + 12x + 8}.

This expression is a perfect cube because:

\dpi{120} \mathrm{\sqrt[3]{\mathrm{x}^3} x}
\dpi{120} \sqrt[3]{8} \sqrt[3]{2^3} 2
\dpi{120} \mathrm{3\cdot x^2\cdot 2 6x^2}
\dpi{120} \mathrm{3\cdot 2^2\cdot x 12x}

Then we can factor the expression, writing it as the cube of the sum of two terms:

\dpi{120} \mathrm{x^3 + 6x^2 + 12x + 8 (x + 2)^3}

Factoring the sum or difference of two cubes

If the expression is a sum or difference of two cubes, we can factor as follows:

\dpi{120} \mathrm{a^3 + b^3 (a+b)\cdot (a^2 - ab+b^2)}
\dpi{120} \mathrm{a^3 - b^3 (a-b)\cdot (a^2 - ab+b^2)}

Example:

factor the expression \dpi{120} \mathrm{x^3 - 64}.

Note that the expression can be written as \dpi{120} \mathrm{x^3 - 4^3}, so it is a difference of two cubes.

Then we can factor the expression as follows:

\dpi{120} \mathrm{x^3 - 64 (x - 4)\cdot (x^2 - 4x+16)}

You may also be interested:

  • algebraic fractions
  • Adding and Subtracting Algebraic Fractions
  • Multiplying and dividing algebraic fractions

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