Mathematics

Polynomials and Rational Expressions

Monomials and polynomials, polynomial operations (evaluation, addition, multiplication, long division, and synthetic division), special products, Binomial Theorem, roots, Remainder and Factor theorems, factoring, and operations with rational expressions.

Monomial: An algebraic expression in which the only operations between the variables are multiplication and powers with natural exponents. Examples:
  • 2x3y2  ;3x4  ;14x32x^3y^2\;; \quad 3x^4\;; \quad \dfrac{1}{4}x^3  are monomials
  • 2x34y2  ;3x−1  ;14x22x^{\frac{3}{4}}y^2\;; \quad 3x^{-1}\;; \quad \dfrac{1}{4x^2}  are NOT monomials
Polynomial: The sum of several monomials. If there are exactly two, it is called a binomial. Examples:
P(x,y)=2x3y2+3x4+14x3Q(x)=x3−2x2+3x+1P(x,y) = 2x^3y^2 + 3x^4 + \dfrac{1}{4}x^3 \qquad Q(x) = x^3 - 2x^2 + 3x + 1
Evaluating a polynomial: The numerical value of a polynomial P(a) is the result of substituting the variable with a. Ex.:
P(x)=3x2−x+1P(x) = 3x^2 - x + 1
P(2)=3⋅22−2+1=11P(2) = 3 \cdot 2^2 - 2 + 1 = 11
Adding/Subtracting polynomials: Combine like terms. Ex.:
(x2+2xy+3xy2+2y2)+(3x2−5xy+2xy2+y3)=\left(x^2 + 2xy + 3xy^2 + 2y^2\right) + \left(3x^2 - 5xy + 2xy^2 + y^3\right) =
=4x2−3xy+5xy2+2y2+y3= 4x^2 - 3xy + 5xy^2 + 2y^2 + y^3
Multiplication:
(x2+2x)⋅(3x2−x+1)=\left(x^2 + 2x\right) \cdot \left(3x^2 - x + 1\right) =
=3x4−x3+x2+6x3−2x2+2x== 3x^4 - x^3 + x^2 + 6x^3 - 2x^2 + 2x =
=3x4+5x3−x2+2x= 3x^4 + 5x^3 - x^2 + 2x
Polynomial Long Division: Both polynomials must be arranged in descending order of degree. The degree of the dividend must be greater than or equal to the degree of the divisor. Ex.:
P(x)Q(x)=4x4−2x3+5x2+1x2−x+1\dfrac{P(x)}{Q(x)} = \dfrac{4x^4 - 2x^3 + 5x^2 + 1}{x^2 - x + 1}
Polynomial long division of 4x⁴−2x³+5x²+1 by x²−x+1, quotient 4x²+2x+3 and reminder x−2
Result:
4x4−2x3+5x2+1x2−x+1=4x2+2x+3+x−2x2−x+1\dfrac{4x^4 - 2x^3 + 5x^2 + 1}{x^2 - x + 1} = 4x^2 + 2x + 3 + \dfrac{x-2}{x^2 - x + 1}
Synthetic Division (Ruffini's Rule): Used when dividing a polynomial by a binomial of the form (x-a). Write the coefficients of the dividend (use 0 for missing terms). Bring down the first coefficient, multiply it by a, add to the next coefficient, and repeat. Ex.:
P(x)Q(x)=4x4−4x3+2x2+1x−2\dfrac{P(x)}{Q(x)} = \dfrac{4x^4 - 4x^3 + 2x^2 + 1}{x - 2}
4−4201288204044102041\begin{array}{r|rrrrr} & 4 & -4 & 2 & 0 & 1 \\ 2 & & 8 & 8 & 20 & 40 \\ \hline & 4 & 4 & 10 & 20 & 41 \end{array}
4x4−4x3+2x2+1x−2=4x3+4x2+10x+20+41x−2\dfrac{4x^4 - 4x^3 + 2x^2 + 1}{x - 2} = 4x^3 + 4x^2 + 10x + 20 + \dfrac{41}{x-2}
(a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2
(2x+3)2=4x2+12x+9(2x+3)^2 = 4x^2 + 12x + 9
(a−b)2=a2−2ab+b2(a-b)^2 = a^2 - 2ab + b^2
(3−3x)2=9−18x+9x2(3-3x)^2 = 9 - 18x + 9x^2
(a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2
(2+x)(2−x)=4−x2(2+x)(2-x) = 4 - x^2
Calculating the power of a binomial
(a+b)n=(n0)anb0+(n1)an−1b1+(n2)an−2b2+⋯+(nn)a0bn=∑k=0n(nk)an−kbk(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 + \dots + \binom{n}{n}a^0 b^n = \sum_{k=0}^n \binom{n}{k} a^{n-k}b^k
Example: Expand (x+2y)3(x+2y)^3
(30)x3(2y)0+(31)x2(2y)1+(32)x1(2y)2+(33)x0(2y)3=\binom{3}{0}x^3(2y)^0 + \binom{3}{1}x^2(2y)^1 + \binom{3}{2}x^1(2y)^2 + \binom{3}{3}x^0(2y)^3 =
=x3+6x2y+12xy2+8y3= x^3 + 6x^2y + 12xy^2 + 8y^3
Pascal's Triangle (or Tartaglia's Triangle).
Allows calculating the binomial coefficients:
n=01n=111n=2121n=31331n=414641…\begin{matrix} \htmlClass{text-primary}{n=0} & & & & & 1 & & & & \\ \htmlClass{text-primary}{n=1} & & & & 1 & & 1 & & & \\ \htmlClass{text-primary}{n=2} & & & 1 & & 2 & & 1 & & \\ \htmlClass{text-primary}{n=3} & & 1 & & 3 & & 3 & & 1 & \\ \htmlClass{text-primary}{n=4} & 1 & & 4 & & 6 & & 4 & & 1 \\ \htmlClass{text-primary}{\dots} & & & & & & & & & \end{matrix}
Root of a polynomial: a is a root of a polynomial if P(a)=0, meaning it is a solution to the equation formed by setting the polynomial equal to 0.
The roots of a polynomial can be found using synthetic division, the quadratic formula, etc.
Example: find the roots of P(x)=2x2−x−1P(x) = 2x^2 - x - 1
Set it equal to 0 and solve:
2x2−x−1=02x^2 - x - 1 = 0
⇒x=1±1−4⋅2⋅(−1)4\Rightarrow \quad x = \dfrac{1 \pm \sqrt{1 - 4 \cdot 2 \cdot (-1)}}{4}
x=1;x=−12x = 1 \quad ; \quad x = -\dfrac{1}{2}
Every polynomial equation of degree n with complex coefficients has exactly n roots in the complex number system (some of which may be repeated or non-real).
The remainder of the division of a polynomial P(x) by a binomial of the form (x-a) is equal to the value of the polynomial evaluated at a, P(a).
Example: find m so that P(x)=2x2−mx−6P(x) = 2x^2 - mx - 6 is divisible by (x-3).
If it is divisible by (x-3), the remainder must be 0. Thus, by the Remainder Theorem:
P(3)=2⋅32−m⋅3−6=0P(3) = 2 \cdot 3^2 - m \cdot 3 - 6 = 0
which gives m = 4
If a is a root of the polynomial P(x), then P(x) is divisible by (x-a), or in other words, (x-a) is a factor of P(x).
Factoring out the greatest common factor (GCF): Ex.:
P(x)=2x3−2x2−4x=2x(x2−x−2)P(x) = 2x^3 - 2x^2 - 4x = 2x\left(x^2 - x - 2\right)
By Synthetic Division: We find the roots using synthetic division. Ex.:
P(x)=2x2−2x−4P(x) = 2x^2 - 2x - 4
2−2−424422022−1−220\begin{array}{r|rrr} & 2 & -2 & -4 \\ 2 & & 4 & 4 \\ \hline & 2 & 2 & 0 \end{array} \qquad \begin{array}{r|rr} & 2 & 2 \\ -1 & & -2 \\ \hline & 2 & 0 \end{array}
The roots are x=2 and x=-1, therefore:
P(x)=2(x−2)(x+1)P(x) = 2(x-2)(x+1)
Using the Quadratic Formula: second-degree polynomials of the form P(x)=Ax2+Bx+CP(x) = Ax^2 + Bx + C can be factored as P(x)=A(x−a)(x−b)P(x) = A(x-a)(x-b), where a and b are the roots (obtained via the quadratic formula), and A is the leading coefficient. Ex.:
P(x)=3x2+12x−15P(x) = 3x^2 + 12x - 15
x=−12±144−4⋅3⋅(−15)6=−12±186x=−5  ;x=1x = \dfrac{-12 \pm \sqrt{144 - 4 \cdot 3 \cdot (-15)}}{6} = \dfrac{-12 \pm 18}{6} \quad x = -5\;; \quad x = 1
P(x)=3(x+5)(x−1)P(x) = 3(x+5)(x-1)
Using Special Products: Ex.:
P(x)=4x2+12x+9=(2x+3)2P(x) = 4x^2 + 12x + 9 = (2x+3)^2
Q(x)=x2−x+14=(x−12)2Q(x) = x^2 - x + \dfrac{1}{4} = \left(x - \dfrac{1}{2}\right)^2
R(x)=9x2y2−4=(3xy−2)(3xy+2)R(x) = 9x^2y^2 - 4 = (3xy-2)(3xy+2)
Simplifying rational expressions: Factor the numerator and the denominator. Common factors cancel out. Ex.:
2x3+2x2−4xx3−x=2x(x−1)(x+2)x(x−1)(x+1)=2(x+2)x+1\dfrac{2x^3 + 2x^2 - 4x}{x^3 - x} = \dfrac{2x(x-1)(x+2)}{x(x-1)(x+1)} = \dfrac{2(x+2)}{x+1}
Adding/subtracting rational expressions: Factor the denominators and find a common denominator. Then operate the expression in the numerator. Simplify the resulting fraction if possible. Ex.:
2x−1+x+1x2−1=2x−1+x+1(x−1)(x+1)=2(x+1)+x+1(x−1)(x+1)\dfrac{2}{x-1} + \dfrac{x+1}{x^2-1} = \dfrac{2}{x-1} + \dfrac{x+1}{(x-1)(x+1)} = \dfrac{2(x+1)+x+1}{(x-1)(x+1)}
=3x+3(x−1)(x+1)=3(x+1)(x−1)(x+1)=3x−1= \dfrac{3x+3}{(x-1)(x+1)} = \dfrac{3(x+1)}{(x-1)(x+1)} = \dfrac{3}{x-1}
Multiplication: Multiply numerators with numerators and denominators with denominators, then simplify. It is recommended to factor first:
2x2+x−2⋅x2−12x+2=2(x−1)(x+2)⋅(x−1)(x+1)2(x+1)\dfrac{2}{x^2+x-2} \cdot \dfrac{x^2-1}{2x+2} = \dfrac{2}{(x-1)(x+2)} \cdot \dfrac{(x-1)(x+1)}{2(x+1)}
=2(x−1)(x+1)(x−1)(x+2)⋅2(x+1)=1x+2= \dfrac{2(x-1)(x+1)}{(x-1)(x+2) \cdot 2(x+1)} = \dfrac{1}{x+2}
Division: Factor numerator and denominator. Multiply crosswise (multiply by the reciprocal) and simplify:
2x+2x2+x−2÷4x+43x−3=2(x+1)(x−1)(x+2)÷4(x+1)3(x−1)\dfrac{2x+2}{x^2+x-2} \div \dfrac{4x+4}{3x-3} = \dfrac{2(x+1)}{(x-1)(x+2)} \div \dfrac{4(x+1)}{3(x-1)}
=2(x+1)⋅3(x−1)(x−1)(x+2)⋅4(x+1)=32(x+2)= \dfrac{2(x+1) \cdot 3(x-1)}{(x-1)(x+2) \cdot 4(x+1)} = \dfrac{3}{2(x+2)}
It is the reverse process of adding rational expressions. It transforms a rational expression into a sum of simpler fractions. The first step is to find the roots and factor the denominator. For each obtained factor, we create corresponding fractions as shown below:
1. For each distinct linear factor (x−a)(x-a), there will be a term of the form Ax−a\dfrac{A}{x-a}
2. For each repeated linear factor (x−b)n(x-b)^n, there will be n terms:
B1x−b+B2(x−b)2+…+Bn(x−b)n\dfrac{B_1}{x-b} + \dfrac{B_2}{(x-b)^2} + \ldots + \dfrac{B_n}{(x-b)^n}
3. For each irreducible quadratic factor (ax2+bx+c)(ax^2+bx+c), there will be a term of the form Cx+Dax2+bx+c\dfrac{Cx+D}{ax^2+bx+c}
Example:
7x+1x3+x2−x−1=7x+1(x−1)(x+1)2=Ax−1+Bx+1+C(x+1)2\dfrac{7x+1}{x^3+x^2-x-1} = \dfrac{7x+1}{(x-1)(x+1)^2} = \dfrac{A}{x-1} + \dfrac{B}{x+1} + \dfrac{C}{(x+1)^2}
To find the unknowns A, B, and C, we set the numerators of both sides equal:
7x+1=A(x+1)2+B(x+1)(x−1)+C(x−1)7x + 1 = A(x+1)^2 + B(x+1)(x-1) + C(x-1)
We substitute as many arbitrary values for x as there are unknowns, to generate a system of equations:
x=1⇒8=4A⇒A=2x=1 \Rightarrow 8 = 4A \Rightarrow A=2
x=−1⇒−6=−2C⇒C=3x=-1 \Rightarrow -6=-2C \Rightarrow C=3
x=0⇒1=A−B−C⇒B=−2x=0 \Rightarrow 1=A-B-C \Rightarrow B=-2
Yielding2x−1−2x+1+3(x+1)2\text{Yielding} \quad \dfrac{2}{x-1} - \dfrac{2}{x+1} + \dfrac{3}{(x+1)^2}