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

### Vocabulary, combing liketerms, multiplyng and factoring polynomials of algebra

 VOCABULARY of algebra •  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 COMBINING LIKE TERMS of algebra •  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 MULTIPLYING & DIVIDING TERMS •  Product rule for exponents: (am)(an) = am+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: (am)n = am·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: am/an = am-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 MULTIPLYING POLYNOMIALS of algebra •  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 FACTORING POLYNOMIALS •  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) FACTORING SPECIAL POLYNOMIALS •  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) SOLVING A FIRST-DEGREE EQUATION WITH ONE VARIABLE •  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 SOLVING A FIRST-DEGREE INEQUALITY WITH ONE VARIABLE •  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  