Quadratic Expressions

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4x^2 + 28x + 49 First square root 4x^2 and 49. This will give you 2x and 7. Next multiply 2x and 7, which is 14x, then multiply 14x by 2, to get a final answer of 28x. This product is equal to the value of "bx". Simply plug in the two squares into the form (*first square* + *second square*)^2 The factored form of the original expression would be (2x+7)^2

Perfect Square Trinomial This technique allows you to get straight to the factored form of a quadratic expression regardless of the value of "a" without having to go through all the steps. This technique can only work when the expression covers certain requirements. 1. First, you must check if the value of "ax^2" and "c" can be square rooted. If both terms are perfect squares, you may move onto next step. 2. Next, check if two times the value of both the perfect squares is equal to the value of "bx". In other words, you must multiply square root of "ax^2" and "c", then multiply that product by 2. If the final answer is equal to "bx", you may proceed to next step. 3. Lastly, you must plug the two perfect squares into the expression (*first square* + *second square*)(*first square* + *second square*) or (*first square* + *second square*)^2

3x^2 + 8x +4 Find two numbers that have a sum of 8, and also have a product of 3 times 4, which is 12. In this case, the numbers would be 6 and 2 since they add up to 8, and have a product of 12. Decompose 8x using these two numbers... 3x^2 +6x+2x+4 You can now group factor... 3x (x+2) + 2 (x+2) You can now binomial common factor... The factored form of the original expression would be (x+2) (3x+2)

x^2 + 7x + 12 You must find two numbers that add up to 7, and must also multiply to give you a product of 12. In this case, the numbers 3 and 4 work. The sum of 3 and 4 is 7, and the product of 3 and 4 is 12. Proceed to plug these numbers into the form stated in previous block... The factored form of the original expression would be (x+3) (x+4)

x^2-100 You would first square root both terms, x^2 and 100 in this case. x-25 Now you would plug in both terms, a being x and b being 25, into the form (a+b) (a-b). The factored form would be... (x+25) (x-25)

3x^2 + 6x + 4x + 8 Group the terms into pairs... (3x^2 + 6x) + (4x + 8) Do monomial common factoring for each of the pairs... 3x (x+2) + 4 (x+2) You will notice that you may sometimes be able to do binomial common factoring after doing monomial common factoring... The factored form of the original equation would be... (x+2) (3x+4)

2x (y+z) + 4 (y+z) The common binomial in this set of terms is (y+z). Place this in a bracket and the rest of the terms in another bracket like such... (y+z) (2x (y+z) + 4 (y+z)) Divide both the terms inside the second bracket by the binomial. (y+z) (2x (y+z)/(y+z) + 4 (y+z)/(y+z)) After factoring out the binomial, you should only have the leftover terms in the second bracket. The factored form of the original equation would be... (y+z) (2x + 4)

2x+4x^2+6x^3 The GCF between 2, 4 and 6 is 2. The common variable with the lowest power between x, x^2 and x^3 is x^1. Therefore, you will divide each of the terms in the set by 2x^1. 2x^1 (2x/2x + 4x^2/2x + 6x^3/2x) *When dividing the terms, you need to keep certain rules in mind such as the quotient rule and the fact that a term divided by itself is 1, not 0.* The factored form of the original example would be... 1+2x+3x^2

(2x+2)^2 "a" represents 2x and "b" represents 2 Substitute into a^2+2ab+b^2 (2x)^2 + 2(2x)(2) +(2)^2 *make sure to raise the entire term to the exponent of two, not just the variable or the coefficient* *when doing the middle term (2ab), you must multiply all the terms, for example, 2(2x)(2) = 2 times 2x times 2* The expanded answer would be: 4x^2 + 8x + 4

(2x-2)(2x+2) "a" = 2x and "b" = 2 Substitute into a^2-b^2 (2x)^2-(2)^2 *make sure to raise the entire term to the exponent of two, not just the variable or the coefficient* The expanded answer would be: 4x^2-4

Pay attention to signs associated with the terms when distributing

When multiplying a variable, "x" for example, by another "x", the product is x^2

When there is a "subtraction" sign before a bracket and no number associated with the sign, you distribute a negative 1 into the bracket

Difference of Squares If you recall from the expanding section, there was a section named "Difference of Squares", just like this section. In the expanding section, we went from an expression in the form of (a+b) (a-b) to the expanded form of a^2-b^2. In the factoring section of the difference of squares, we are simply working backwards; from a^2-b^2 to (a+b) (a-b). In order to do so, you must simply square root both terms, a and b, and proceed to plug them into the form (a+b) (a-b).

Factoring Quadratic Expression (a≠1) When you have a quadratic expression when a is not equal to 1, you must take other steps in order to factor the expression. In fact, this type of expression requires steps that incorporate many of the other types of factoring. 1. First, you must find two numbers that add up to the value of b. Those same two numbers must give you a product equal to the value of ac ("a" value multiplied by "c" value). 2. After you have got the two numbers, you must decompose the middle term, "bx", using the two numbers. Decompose simply means to break apart. For example, if "bx" is 6x, and your two numbers are 3 and 3, the middle term decomposed would become 3x+3x. 3. Once you have decomposed the middle term, you will be able to do group factoring. 4. You may also find that after group factoring, you will be able to do binomial common factoring.

Binomial Common Factoring Similar to monomial common factoring, binomial common factoring is where you factor out a common binomial out of a set of terms. You essentially divide the set of terms by a binomial which all of the terms share. 1. Once you have identified the binomial, place the binomial in a bracket. Place the set of terms you are factoring in another bracket right beside the first bracket. 2. Proceed to divide the set of terms by the binomial. 3. After you divide the terms, you should only have singular terms left inside the second bracket.

Difference of Squares When you come across a set of brackets in the form of (a-b)(a+b), instead of using the FOIL technique, you can get straight to the expanded answer by subbing the two terms into the expression : a^2 - b^2 "a" represents the first term, and "b" represents the second term.

Perfect Square Trinomials When you come across a set of identical brackets/a bracket squared ((a+b)^2 or (a-b)^2)), you can get straight to the expanded answer by subbing the two terms into the expression: a^2+2ab+b^2 or a^2-2ab+b^2 "a" represents first term in bracket, and "b" represents second term.

Factoring Quadratic Expressions (a=1) Factoring a quadratic expression requires a few more steps in order to achieve the factored form. Quadratic expression is written in the form of ax^2+bx+c. In this specific type of quadratic expression, "a" is equal to 1, in other words, "a" does not necessarily exist or at least it's not written. 1. When you are given the quadratic expression, first check to see if you are able to monomial common factor out a term. 2. Next, you must find two numbers that give you the SUM of "b". Those same two numbers must also give you the PRODUCT of "c". 3. Once you have gotten your two numbers, simply plug them into the into an expression in the form of (x+*first number*) (x+*second number*)

Factor by Grouping Factoring by grouping is essentially an extension of monomial common factoring and may sometimes incorporate binomial common factoring as well. When you are faced with a set of terms (often times more than 4 terms present) all of which seem like they do not share a GCF or common variable, you can simply group the terms that have similarities into pairs and then proceed to monomial common factor in each of the pairs.

Monomial Common Factoring Monomial common factoring is exactly what the name suggests; you factor out a monomial out of a set of terms. Factoring is essentially dividing the set of terms by the term(s) that you are factoring out. You must follow a few steps in order to monomial common factor correctly: 1. Find the greatest common factor between the coefficients of all the terms in the set you are about to factor. *Often times, you might find that there are no GCF's between the terms. When this happens, keep in mind that the GCF is in fact 1. From here, you can move onto next step. 2. Look for a variable which ALL terms share. If they all share the same variable but with different exponents, take the variable with the lowest power. For example: x+x^2+x^4 You would take x^1 since it is the variable with the lowest power. 3. Write down the GCF and the lowest common variable between the terms, then proceed to write the set of terms you are factoring directly beside the GCF & variable in brackets. 4. Divide all the terms inside the bracket by the constructed term which you place outside of the bracket.

Foil (Multiplying Polynomials) Multiplying polynomials is similar to the distributive Property but is instead used when there are two sets of brackets directly beside one another. Both brackets can only contain two terms. FOIL is an acronym which stands for: First - multiply first terms in both brackets Outer - multiply outer terms in brackets Inner - multiply inner terms in brackets Last - multiply last terms in brackets

Distributive Property When a term exists before a bracket, you must distribute the outer term into all the terms inside the bracket through multiplication. This will allow you to expand the expression/equation by getting rid of the brackets. For example: 2x (2+4-3) expanded is 4x+8x-6x

Quadratic Expressions

Factoring

Expanding