Int Alg TS04 Review

Int. Alg. Notes Test #4 Review Page 1 of 11

Section 5.1: Adding and Subtracting Polynomials A monomial in one variable is the product of a constant and a variable raised to a nonnegative integer power.  General form of a monomial in one variable: axk.  The constant a is called the coefficient and k is the degree of the monomial.  Examples of monomials in one variable:

2x2 -3y5 m6 4z -7

A binomial is the sum of two monomials, like 3p2 – 6p A trinomial is the sum of three monomials, like p2 – 6p + 9 A polynomial is a monomial or the sum of monomials, like 4p3 + 2p2 + 7p – 12  A polynomial is in standard form when it is written with the terms in descending order of degree.  The degree of the polynomial is the highest degree of any of its terms.

A monomial in many variables is the product of a constant and two or more variables raised to nonnegative integer powers.  General form of a monomial (in two variables): axmyn  The constant a is called the coefficient and m + n is the degree of the monomial.

To add polynomials, combine like terms. To subtract polynomials, combine like terms.

A polynomial function is a function whose rule is a polynomial.  The domain of a polynomial function is all real numbers.  The degree of a polynomial function is the value of the largest exponent on the variable.  A polynomial function of degree one or zero is called a linear function, like f( x) =0.5 x - 1, or f( x) = 3.

 A polynomial function of degree two is called a quadratic function, like f( x) = x2 -2 x + 9 .

 A polynomial function of degree three is called a cubic function, like f( x) =2 x3 + 7 x 2 + 5 x - 3.

Algebra is: the study of how to perform multi-step arithmetic calculations more efficiently, and the study of how to find the correct number to put into a multi-step calculation to get a desired answer. Int. Alg. Notes Test #4 Review Page 2 of 11

If f and g are two functions, then  The new function that can be made by adding them together is called f + g: (f + g)(x) = f (x) + g(x). The new function that can be made by subtracting them is called f – g: (f – g)(x) = f (x) – g(x).

Section 5.2: Multiplying Polynomials

Skill #1: Multiplying Monomials When multiplying monomials:  Multiply the coefficients to get a single new coefficient.  Add exponents of common variables to get simplified variable factors. Example:

3 2 4骣 3 2 骣 4 (2a b)(- 6 a b) =( 2( - 6))鬃琪 a a 鬃琪 b b 桫a鬃 a a 鬃 a a 桫 b 鬃 b b 鬃 b b =( -12)(a3+ 2)( b 1 + 4 ) = -12a5 b 5

Skill #2: Multiplying a Monomial and a Polynomial When multiplying a monomial and a polynomial, use the extended form of the distributive property: Extended form of the Distributive Property:

a( b1+ b 2 + b 3 + + bn) = ab 1 + ab 2 + ab 3 +  + ab n Example: 2x2( x 2+ 3 x - 5) = 2 x 2� x 2 � 2 x 2 � 3 x 2 x 2 ( 5) =2x4 + 6 x 3 - 10 x 2

Skill #3: Multiplying a Binomial and a Binomial When multiplying a binomial and a binomial, you can use one of four techniques: 1. Distributive Property Technique Example: (3x+ 4)( 2 x - 9) =( 3 x + 4)� 2 x + ( 3 x � 4) ( 9) =3x� 2 x � 4 � 2 x + 3 x �( 9) 4( 9) =6x2 + 8 x - 27 x - 36 =6x2 - 19 x - 36 2. Vertical Multiplication Technique Example: (3x+ 4)( 2 x - 9) = 3x + 4 � 2x 9 -27x - 36 6x2 + 8 x 6x2 - 19 x - 36 3. Table Multiplication Technique (not in book) Example: To Calculate 3429: To Calculate (3x+ 4)( 2 x - 9) :

 30 4  3x 4 20 600 80 2x 6x2 8x 9 270 36 -9 -27x -36

= 600 + 80 + 270 + 36 = 6x2 + 8x – 27x – 36 = 986 = 6x2 – 19x – 36

4. FOIL Technique (ONLY works for binomials) FOIL  First, Outside, Inside, Last

Skill #4: Multiplying a Polynomial and a Polynomial When multiplying a binomial and a binomial, you can use one of two techniques: 1. Distributive Property Technique Example: (2x+ 3)( x2 + 5 x - 2) =( 2 x + 3) +( 2 x + 3)�( 5 x) + ( 2 x � 3) ( 2) =2x� x2 � 3 � x 2 � 2 x � 5 x + 3 � 5 x 2 x ( 2) 3( 2) =2x3 + 3 x 2 + 10 x 2 + 15 x - 4 x - 6 =2x3 + 13 x 2 + 11 x - 6 2. Table Multiplication Technique (not in book) Example:  x2 5x -2 2x 2x3 10x2 -4x 3 3x2 15x -6 Notice that like terms line up on the diagonals…

Skill #5: Recognizing Special Binomial Products When multiplying a pair of “conjugate” binomials, or when multiplying a binomial with itself, you can use the following formulas: 1. The Product of Conjugate Binomials, or, a Difference of Two Squares (A + B)(A – B) = A2 – B2

2. The Square of a Binomial, or, a Perfect Square Trinomial (A + B)2 = A2 + 2AB + B2 (A – B)2 = A2 – 2AB + B2

If f and g are two functions, then  The new function that can be made by multiplying them together is called f  g: (f  g)(x) = f (x)  g(x).

Section 5.3: Dividing Polynomials; Synthetic Division Dividing a polynomial by a monomial: Divide the monomial into each term of the polynomial, and cancel when possible.

Dividing a polynomial by a polynomial using long division: Long division of polynomials is a lot like long division of numbers: a. Arrange divisor and dividend around the dividing symbol, and be sure to write them in descending order of powers with all terms explicitly stated (even the terms with zero coefficients). b. Divide leading terms, then multiply and subtract. c. Repeat until a remainder of order less than the divisor is obtained. Dividend Remainder =Quotient + Divisor Divisor Example: Compute (5x2 + 7x +9) ÷ (x + 6)

Dividing a polynomial by a binomial using synthetic division: THIS IS A SHORTCUT THAT ONLY WORKS WHEN THE DIVISOR IS A LINEAR BINOMIAL (I.E., THE DIVISOR IS x – c) !!!

Synthetic division is a shorthand way to divide a polynomial by the linear factor x – c: a. Write c outside the division bar and the coefficients of the dividend inside the bar. b. Bring the leading coefficient of the dividend straight down. c. Compute c times the number in the bottom row, and write the answer in the middle row to the right. d. Add and repeat until all coefficients are used up. Example: Compute (2x3 – 3x2 – 4x + 11) ÷ (x – 2) using synthetic division.

Definition: Quotient of Functions f If f and g are two functions, then the new function that can be made by taking their quotient is called , and is g 骣f f( x) defined as: 琪 ( x) = , provided g(x)  0. 桫g g( x)

The Remainder Theorem If the polynomial P(x) is divided by x – c, then the remainder is the value P(c). This is because when we divide a polynomial by x – c, the remainder must be of degree less than x – c, which means the remainder has degree of zero, which is just a numeric constant R. So:

The Factor Theorem If P(x) is a polynomial function, then x – c is a factor of P(x) if and only if R = P(c) = 0. In other words, if P(c) = 0, then P(x) can be written as P(x) = (x – c)(Quotient(x)).

(This can be used to see if a divisor divides evenly into a dividend quickly.)

Section 5.4: Greatest Common Factor; Factoring by Grouping In algebra, when you are asked to factor a polynomial, you are supposed to find other polynomials whose product is the polynomial you are factoring. Examples:  Factor x2 +2 x + 1: x2 +2 x + 1 =( x + 1)( x + 1)  Factor x3+3 x 2 + 3 x + 1: x3+3 x 2 + 3 x + 1 =( x + 1)( x + 1)( x + 1)  Factor x3 -1: x3-1 =( x - 1)( x 2 + x + 1)  The above examples are prime factorizations because each of the factors are prime polynomials. We say that the polynomials above have been factored completely.

In algebra, a polynomial with integer coefficients is prime when it can not be written as the product of two other polynomials with integer coefficients (other than itself and 1). Examples:  7 is prime  ( x +1) is prime  ( x2 + x +1) is also prime

Factoring the greatest common factor: 1. Identify the greatest common factor (GCF) in each term. 2. Rewrite each term as the product of the GCF and the remaining factor. 3. Use the Distributive Property to factor out the GCF. 4. Use the Distributive Property to verify that the factorization is correct. Example: 10x2 y 2+ 15 xy 3 - 25 x 3 y 4 = 5 xy 2� 2 x � 5 xy 3 y 5 xy 5 x 2 y 2 =5xy�( 2 x - 3 y 5 x2 y 2 ) Factoring by grouping: 1. Group terms that have a common factor. You may need to rearrange the terms. 2. In each grouping, factor out the common factor. 3. Factor out the common factor that remains. 4. Check your work.   . Example: 5y- 5 z + ay - az =( 5 y - 5 z) +( ay - az) =5( y - z) + a( y - z) =(5 +a)( y - z)

Section 5.5: Factoring Trinomials Steps for Factoring Quadratic Trinomials of the Form x2 + bx + c:  Find factors of c that sum to b… o List all factors of b o Add up each pair of factors; choose the pair of factors that add up to b  Supposing that the factors are m and n (i.e., mn = c), write the trinomial in factored form as: x2 + bx + c = (x + m)(x + n)  Check your work by multiplying out your factors.

Example:  Factor x2 – 4x – 12  All the factors of -12 are: (+1)(-12), (-1)(+12), (+2)(-6), (-2)(+6), (+3)(-4), (-3)(+4)  The factors that add up to –4 are: (+2) and (-6)   x2 – 4x – 12 = (x+ 2)(x – 6)

Steps for Factoring Quadratic Trinomials of the Form ax2 + bx + c:  Find factors of ac that sum to b… o List all factors of ac o Add up each pair of factors; choose the pair of factors that add up to b  Supposing that the factors are m and n (i.e., mn = ac and m + n = b), re-write the trinomial with the linear term broken up into a sum using m and n: ax2 + bx + c = ax2 + mx + nx+ c  Take your re-written polynomial and factor it by grouping.  Check your work by multiplying out your factors.

Example:  Factor 2x2 – 9x – 18  Multiply leading term coefficient and constant term: (2)(-18) = -36  All the factors of -36 are: (+1)(-36), (-1)(+36), (+2)(-18), (-2)(+18), (+3)(-12), (-3)(+12), (+4)(-9), (-4)(+9), (+6)(-6)  The factors that add up to -9 are: (+3) and (-12)  2x2 – 9x – 18 = 2x2 + 3x – 12x – 18 = x(2x + 3) – 6(2x + 3) = (x – 6) (2x + 3)

Section 5.6: Factoring Special Products Product of conjugate binomials, or a Difference of Squares A2 – B2 = (A + B)(A – B)

Square of a binomial, or a Perfect Square Trinomial A2 + 2AB + B2 = (A + B)2 A2 – 2AB + B2 = (A – B)2

Factoring a Sum of Two Cubes A3+ B 3 =( A + B)( A 2 - AB + B 2 )

Factoring a Difference of Two Cubes A3- B 3 =( A - B)( A 2 + AB + B 2 )

Section 5.7: Factoring: A General Strategy

Section 5.8: Polynomial Equations The Zero-Product Property If the product of two numbers is zero, then at least one of the numbers is zero. That is, if ab = 0, then a = 0 or b = 0, or both a and b are 0.

Steps for solving any polynomial equation (degree 2 or higher): 1. Expand the polynomial equation (if needed), and collect all terms on one side; combine like terms. 2. Factor the polynomial on the one side. 3. Set each factor from step 2 equal to zero (this is justified by the zero-product property rule). 4. Solve each first degree equation for the variable. 5. Check answers in original equation.