Directions: Fill in the boxes with any numbers that make the equation true.
Source: Open Middle
|
The term factor is used a lot in mathematics, so it is important to understand exactly what is meant when we say to factor an expression.
Arithmetically, to factor means to find two numbers, that when multiplied together, will equal a given number. For example, the numbers \(3\) and\(5\) are one set of factors of the number \(15\).
In algebra, when we ask you to factor, we are asking you to find the expressions, that when multiplied together, will equal a larger expression. It is important to know how to factor expressions because it is a quick and easy method that can be used to solve quadratic equations written in standard form.
Before we begin to factor, it is important to review a few terms that will be used in factoring.
We will be factoring binomials and trinomials in this unit, using several types of factoring methods and patterns. If a polynomial cannot be factored, we say that it is prime.
Greatest Common Factoring (GCF)
For any factoring problem, you should always look to see if a greatest common factor exists. The greatest common factor is the monomial that will divide each term of the expression. It will be the largest number that will divide out of each term and the variable raised to the highest power that will divide out of each term.
Example 1: Factor the expression completely: \(27x^3-6x^2+9x\).
The largest number that will divide \(27\), \(6\) and \(9\) is \(3\). Each term also has an \(x\) that can be divided out. So the GCF is \(3x\). Write the factor as the product of the GCF (which is a monomial) and the trinomial that is left over after dividing each term by \(3x.\)
The expression \(27x^3-6x^2+9x\) factors into \(3x\left(9x^2-2x+3\right)\).
This can be checked by distributing the \(3x\) to each term in the trinomial. It is also important to factor the trinomial if possible.
Example 2: Factor the expression completely: \(15x^2+45x\).
Arithmetically, to factor means to find two numbers, that when multiplied together, will equal a given number. For example, the numbers \(3\) and\(5\) are one set of factors of the number \(15\).
In algebra, when we ask you to factor, we are asking you to find the expressions, that when multiplied together, will equal a larger expression. It is important to know how to factor expressions because it is a quick and easy method that can be used to solve quadratic equations written in standard form.
Before we begin to factor, it is important to review a few terms that will be used in factoring.
- A monomial is an expression that is either a number, a variable, or the product of a number and one or more variables (including powers). For example: \(2\), \(3x\), \(-4x^2\).
- A binomial is an expression that is a sum or difference of two monomials. For example: \(x-7\), \(2x^2+3\).
- A trinomial is the sum or difference of three monomials. For example: \(x^2-4x+1\).
We will be factoring binomials and trinomials in this unit, using several types of factoring methods and patterns. If a polynomial cannot be factored, we say that it is prime.
Greatest Common Factoring (GCF)
For any factoring problem, you should always look to see if a greatest common factor exists. The greatest common factor is the monomial that will divide each term of the expression. It will be the largest number that will divide out of each term and the variable raised to the highest power that will divide out of each term.
Example 1: Factor the expression completely: \(27x^3-6x^2+9x\).
The largest number that will divide \(27\), \(6\) and \(9\) is \(3\). Each term also has an \(x\) that can be divided out. So the GCF is \(3x\). Write the factor as the product of the GCF (which is a monomial) and the trinomial that is left over after dividing each term by \(3x.\)
The expression \(27x^3-6x^2+9x\) factors into \(3x\left(9x^2-2x+3\right)\).
This can be checked by distributing the \(3x\) to each term in the trinomial. It is also important to factor the trinomial if possible.
Example 2: Factor the expression completely: \(15x^2+45x\).
Factoring trinomials \(ax^2+bx+c\), where \(a=1\)
- When factoring a trinomial of this form, there will be two binomial factors.
- To factor trinomials whose leading coefficient is \(1\), find the factors of the constant, \(c\), that add \(b\), the coefficient of \(x\).
Example 3: Factor the expression completely: \(x^2-2x-8\).
To factor this expression, find the factors of \(-8\) that add to \(-2\). To illustrate, all of the factors of the constant and their sums are shown.
We write the binomial factors as \(\left(x+2\right)\left(x-4\right)\ or\ \left(x-4\right)\left(x+2\right)\). Since multiplication is commutative, the order the factors are written does not matter.
Example 4: Factor the expression completely: \(x^2+15x+50\).
The two numbers that multiply to \(50\) and add to \(15\) are \(5\) and \(10\). So the correct factorization of \(x^2+15x+50\) is \(\left(x+5\right)\left(x+10\right)\)
Example 5: Factor the expressions completely:
a. \(x^2-11x+28\)
|
b. \(2x^2+12x-54\)
|
c. \(x^3-4x^2-5x\)
|
Factoring trinomials \(ax^2+bx+c\), where \(a\ne1\)
- When factoring a trinomial of this form, there will be two binomial factors.
- To factor trinomials where \(a\ne1\), find all the positive factors of the leading coefficient, \(a\), and all the factors of the constant, \(c\), so that when placed in binomials the product of the outer terms plus the product of the inner terms will add to \(b\), the coefficient of \(x\). This is trial and error, but gets easier with practice.
Example 6: Factor the expression completely: \(5x^2-17x+6\).
The factors of the leading coefficient are \(1\) and \(5\). The factors of the constant are \(1\) and \(6\), \(-1\) and \(-6\), \(2\) and \(3\), \(-2\) and \(-3\). Since \(b\) is negative, you can eliminate the positive factors (because two positive numbers will never add to a negative number). So we are only going to use \(-1\) and \(-6\), and \(-2\) and \(-3\). Let’s place these in a table to organize the information.
The correct factorization of \(5x^2-17x+6\) is \(\left(x-3\right)\left(5x-2\right)\)
Example 7: Factor the expression completely: \(4x^2+12x+5\).
The factors of the leading coefficient are \(1\) and \(4\), or \(2\) and \(2\).
The factors of the constant are \(1\) and \(5\) (we do not need \(-1\) and \(-5\) because the middle term is positive.
Example 7: Factor the expression completely: \(4x^2+12x+5\).
The factors of the leading coefficient are \(1\) and \(4\), or \(2\) and \(2\).
The factors of the constant are \(1\) and \(5\) (we do not need \(-1\) and \(-5\) because the middle term is positive.
The correct factorization of \(4x^2+12x+5\) is \(\left(2x+1\right)\left(2x+5\right)\)
Example 8: Factor the expressions completely:
a. \(8x^2-2x-15\)
|
b. \(6x^2-13x-5\)
|
c. \(4x^2+27x+18\)
|
Perfect Square Trinomials
If the factors of a and c in a trinomial are both perfect squares, check to see if the trinomial fits one of the special factoring patterns shown below:
\(a^2+2ab+b^2=\left(a+b\right)\left(a+b\right)=\left(a+b\right)^2\)
\(a^2-2ab+b^2=\left(a-b\right)\left(a-b\right)=\left(a-b\right)^2\)
If the factors of a and c in a trinomial are both perfect squares, check to see if the trinomial fits one of the special factoring patterns shown below:
\(a^2+2ab+b^2=\left(a+b\right)\left(a+b\right)=\left(a+b\right)^2\)
\(a^2-2ab+b^2=\left(a-b\right)\left(a-b\right)=\left(a-b\right)^2\)
Example 9: Factor the expression completely: \(x^2-8x+16\).
The leading coefficient, \(1\), and the constant, \(16\), are both perfect squares. Since the middle term is twice the product of \(1\) and \(4\) (the square roots of the leading coefficient and the constant), this trinomial is a perfect square trinomial.
The correct factorization of \(x^2-8x+16\) is \(\left(x-4\right)^2\)
Example 10: Factor the expression completely: \(9x^2+30x+25\).
The leading coefficient and the constant are both perfect squares. Since the middle term is twice the product of \(3\) and \(5\), this is a perfect square trinomial.
The correct factorization of \(9x^2+30x+25\) is \(\left(3x+5\right)^2\)
Keep in mind, if you do not see this pattern, you can factor using methods for trinomials with \(a = 1\) or \(a\ne1\).
Difference of Squares
The difference of squares factoring pattern can be used if the two terms in a binomial are both perfect squares and are joined by subtraction. The difference of squares factoring pattern is:
\(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
The difference of squares factoring pattern can be used if the two terms in a binomial are both perfect squares and are joined by subtraction. The difference of squares factoring pattern is:
\(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
Example 11: Factor the expression completely: \(x^2-100\).
Since the terms in the binomial are both perfect squares joined by subtraction, this fits the pattern. The factors are \(\left(x-10\right)\left(x+10\right)\)
(Basically, we are looking for two numbers that multiply to \(-100\) that add to zero)
Example 12: Factor the expressions completely:
a. \(25x^2-36\)
|
b. \(2x^2-50\)
|