Algebra - vocabulary, combing liketerms, multiplyng and factoring polynomials, table

•  Variables are letters or symbols used to represent unknown numbers

•  Constants are specific numbers that are not multiplied by any variables

•  Coefficients are numbers that are multiplied by one or more variables

EX: -4xy has a coefficient of -4; 9m³ has a coefficient of 9; x has an implied coefficient of 1

•  Terms are constants or variable expressions EX: 3a, -5c⁴d, 25mp³r⁵, and 7 are all terms

•  Like or similar terms are terms that have the same variables to the same degree or exponent value; coefficients may or may not be equal

EX: 3m² and 7m² are like terms because they both have the same variable to the same power or exponent value; -15a⁶b and 6a⁶b are like terms; 2X⁴ and 6x³ are not like terms because, although they have the same variable (x), it is to the power of 4 in one term and to the power of 3 in the other

•  Algebraic expressions are terms that are connected by either addition or subtraction EX: 2s + 4a² - 5 is an algebraic expression with three terms: 2s, 4a², and -5

•  Algebraic equations are statements of equality between at least two terms

EX: 4z = 28 is an algebraic equation; 3(a - 4) + 6a = 10 - a is also an algebraic equation; note that both statements have equal signs in them

•  Algebraic inequalities are statements that have either > or < between at least two terms

EX: 50 < -2x is an algebraic inequality; 3(2n + 7) > -10 is an algebraic inequality

•  You can add or subtract coefficients of like terms to simplify an algebraic expression; when doing so, the value of the exponent does not change

EX: 3a + 7a = 10a; 9d² - 6d² = 3d²

EX: Axy³ and -7y³x are like terms, even though the x and y³ are not in the same order, and may be combined in this manner: 4xy³ + -7y³x = -3xy³; -15a²bc and 3bca⁵ are not like terms because the exponents of the a are not the same in both terms, so the coefficients may not be added or subtracted

•  Product rule for exponents: (a^m)(aⁿ) = a^m+n; that is, when multiplying terms with the same base (a in this case), add the exponents

Note: Any terms may be multiplied, not just like terms

EX: (n⁸)(n⁵) = n¹³

•  When terms include coefficients, multiply the coefficients and follow the product rule for exponents

EX: (4a⁴c)(-12a²b³c) = -48a⁶b³c²; note that 4 times -12 became -48, a⁴ times a² became a⁶, c times c became c², and the b³ was written to indicate multiplication by b, but the exponent did not change on the b because there was only one b in the expression

•  Power rule for exponents: (a^m)ⁿ = a^m·n; that is, when raising a term to a power (a in this case), multiply the exponents

EX: (c³)⁴ = c¹²

•  When terms include coefficients, raise the coefficients to the power and follow the power rule for exponents

EX: (2a²b⁵c)⁵ = 32a¹⁰b²⁵c⁵

•  Quotient rule for exponents: a^m/aⁿ = a^m-n; that is, when dividing a term to a power by a term to a power with the same base (a in this case), subtract the exponents

EX: p⁸ ÷ p² = p⁶

•  When terms include coefficients, divide the coefficients and follow the quotient rule for exponents

EX: -27a⁶b ÷ 9a⁵ = -3ab

•  Rule for zero exponents: a⁰ = 1; that is, any nonzero base to a 0 power (a in this case) equals 1

EX: x⁰ = 1

•  Rule for negative exponents: a^-n = 1/aⁿ; that is, any nonzero base to a negative power (a in this case) equals its reciprocal raised to the positive power

EX: x^-3 = 1/x³

•  When a polynomial is multiplied by a monomial, use the distributive property: a(c + d) = ac + ad

EX: 4x³(2xy + y²) = 8x⁴y + 4x³y²

•  When multiplying a polynomial by a polynomial, be sure to multiply each term in the first polynomial by each term in the second polynomial, and then combine like terms; this is often referred to as using the FOIL method for products of binomials; essentially, when multiplying two binomials, multiply each First term, each Outside term, each Inside term, and each Last term; then combine like terms: (a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd

EX: (2x + y)(3x - 5y) = 2x(3x - 5y) + y(3x - 5y) = 6x² - 10xy + 3xy - 5y² = 6x² - 7xy - 5y²

•  When you factor a polynomial, you are finding what two polynomials multiply together to result in a product of the original polynomial; there are a variety of factoring strategies; when factoring a trinomial (a polynomial with three terms), one common method is to use trial and error by making an educated guess of the first factor in each binomial and an educated guess of the second factor in each binomial, then multiplying the polynomials and adjusting if needed

EX: Factor  x2 + 7x + 12; try (x + 6)(x + 2); the product is x² + 8x + 12, so adjust the second term in each binomial to be other factors of 12; try (x + 4)(x + 3); the product is x² + 7x + 12; the factors are (x + 4) and (x + 3)

EX: Factor 6x² - 5x - 6; try (2x + 3)(3x - 2); the product is 6x² + 5x - 6, so adjust the signs in each binomial; try (2x - 3)(3x + 2); the product is 6x² - 5x - 6; the factors are (2x - 3) and (3x + 2)

•  To factor a difference of squares, use the rule a² - b² = (a + b)(a - b)

EX: Factor x² - 25

x² and 25 are square numbers; (x)² = x² and (5)² = 25; x² - 25 = (x + 5)(x - 5)

•  To factor a difference of cubes, use the rule a³ - b³ = (a - b)(a² + ab + b²)

EX: Factor 8x³ - 64

8x³ and 64 are cube numbers; (2x)³ = 8x³ and (4)³ = 64; 8x³ - 64 = (2x - 4)[(2x)² + (2x)(4) + (4)²] = (2x - 4)(4x² + 8x + 16)

•  To factor a sum of cubes, use the rule a³ + b³ = (a + b)(a2 - ab + b2)

EX: Factor 27x³ + 125

27x³ and 125 are cube numbers; (3x)³ = 27x³ and (5)³ = 125; 27x³ + 125 = (3x + 5)[(3x)² -(3x)(5) + (5)²] = (3x + 5)(9x² - 15x + 25)

•  Eliminate any decimal fractions by using the multiplication property of equality

EX: 1/2·(3a + 5) = 2/3·(7a - 5) + 9 would be multiplied on both sides of the equals sign by the lowest common denominator of  1/2  and  2/3  which is 6; the result would be 3(3a + 5) = 4(7a - 5) + 54; note that only 1/2, 2/3, and 9 were multiplied by 6 and not the contents of the parentheses; the parentheses will be handled in the next step, when the distributive property is used

•  Use the distributive property to remove any parentheses, if there are any

EX: 3(3a + 5) = 4(7a - 5) + 54 becomes 9a + 15 = 28a - 20 + 54

•  Combine any like terms that are on the same side of the equals sign

EX: 9a + 15 = 28a - 20 + 54 becomes 9a + 15 = 28a + 34 because the only like terms on the same side of the equals sign were -20 and +54

•  Use the addition property of equality or subtraction property of equality to add or subtract the same terms on both sides of the equals sign; this may be done more than once; the objective is to get all terms with the same variable on one side of the equals sign and all terms without the variable on the other side of the equals sign

EX: 9a + 15 = 28a + 34 becomes 9a + 15 - 28a - 15 = 28a + 34 - 28a - 15; note that both -28a and -15 were added to both sides of the equals sign at the same time; this results in -19a = 19 after like terms are added or subtracted

•  Use the multiplication property of equality or division property of equality to make the coefficient of the variable 1

EX: -19a = 19 would be multiplied on both sides by -1/19 (or divided by -19), so the equation would become -19a(-1/19) = 19a(-1/19)  or a = -1

•  Check the answer by substituting it for the variable in the original equation to see if it makes the original equation true

•  Follow the same steps for solving a first-degree equality as described above, except for one step in the process; this exception follows:

•  Exception: When applying the multiplication property, the inequality sign must reverse if you multiplied by a negative number

EX: In 4m > -48, you need to multiply both sides of the > symbol by 1/4; therefore,

4m(1/4) > -48(1/4); this results in m > -12; note that the > did not reverse because you multiplied by a positive 1/4; however, in -5x > 65, you need to multiply both sides by -1/5; therefore, -5x(-1/5) < 65(-1/5); this results in x < -13; note that the > did reverse and become < because you multiplied by a negative number, -1/5

•  Check the solution by substituting some numerical value for the variable in the original inequality