How To Help With Polynomials

Polynomials have more than one term. They contain constants, variables and exponents. The constants, called coefficients, are the multiplicands of the variable, a letter that represents an unknown mathematical value within the polynomial. Both the coefficients and the variables may have exponents, which represent the number of times to multiply the term by itself. You can use polynomials in algebraic equations to help find the x-intercepts of graphs and in a number of mathematical problems to find values of specific terms.

Step 1

Examine the expression -9x^6 – 3. To find the degree of a polynomial, find the highest exponent. In the expression -9x^6 – 3, the variable is x and the highest power is 6.

Step 2

Examine the expression 8x^9 – 7x^3 + 2x^2 – 9. In this case, the variable x appears three times in the polynomial, each time with a different exponent. The highest variable is 9.

Step 3

Examine the expression 4x^3y^2 – 3x^2y^4. This polynomial has two variables, y and x, and both are raised to different powers in each term. To find the degree, add the exponents on the variables. X has a power of 3 and 2, 3 + 2 = 5, and y has a power of 2 and 4, 2 + 4 = 6. The degree of the polynomial is 6.

Step 1

Simplify the polynomials with addition: (4x^2 – 3x + 2) + 6x^2 + 7x – 5). Combine like terms to simplify added polynomials: (4x^2 + 6x^2) + (-3x + 7x) + (2 – 5) = 10x^2 + 4x – 3.

Step 2

Simplify the polynomials with subtraction: (5x^2 – 3x + 2) – (2x^2 – 7x – 3). First, distribute, or multiply the negative sign: (5x^2 – 3x + 2) – 1(2x^2 – 7x – 3) = 5x^2 – 3x + 2 – -2x^2 + 7x + 3. Combine like terms: (5x^2 – 2x^2) + (-3x + 7x) + (2 + 3) = 3x^2 + 4x + 5.

Step 3

Simplify the polynomials with multiplication: 4x(3x^2 + 2). Distribute the term 4x by multiplying it to each of the terms within parentheses: (4x)(3x^2) + (4x)(2) = 12x^3 + 8x.

Step 1

Examine the polynomial 15x^2 – 10x. Before beginning any factorization, always look for the greatest common factor. In this case, the GCF is 5x. Pull the GCF out, divide the terms and write the remainder in parentheses: 5x(3x – 2).

Step 2

Examine the expression 18x^3 – 27x^2 + 8x – 12. Reorder the polynomials to factor one set of binomials at a time: (18x^3 – 27x^2) + (8x – 12). This is called grouping. Pull out the GCF of each binomial, divide and write the remainders in parentheses: 9x^2(2x – 3) + 4(2x – 3). The parentheses must match for group factorization to work. Finish factoring by writing the terms in parentheses: (2x – 3)(9x^2 + 4).

Step 3

Factor the trinomial x^2 – 22x + 121. Here there is no GCF to pull out. Instead, find the square roots of the first and last terms, which in this case are x and 11. When setting up the parenthetical terms, remember the middle term will be the sum of the products of the first and last terms.

Step 4

Write the square root binomials in parenthetical notation: (x – 11)(x – 11). Redistribute to check the work. The first terms, (x)(x) = x^2, (x)(-11) = -11x, (-11)(x) = -11x and (-11)(-11) = 121. Combine like terms, (-11x) + (-11x) = -22x, and simplify: x^2 – 22x + 121. Since the polynomial matches the original, the process is correct.

Step 1

Examine the polynomial equation 4x^3 + 6x^2 – 40x = 0. This is the zero product property, which allows the terms to move to the other side of the equation to find the value(s) of x.

Step 2

Factor out the GCF, 2x(2x^2 + 3x – 20) = 0. Factor out the parenthetical trinomial, 2x(2x – 5)(x + 4) = 0.

Step 3

Set the first term to equal zero; 2x = 0. Divide both sides of the equation by 2 to get x by itself, 2x ÷ 2 = 0 ÷ 2 = x = 0. The first solution is x = 0.

Step 4

Set the second term to equal zero; 2x^2 – 5 = 0. Add 5 to both sides of the equation: 2x^2 – 5 + 5 = 0 + 5, then simplify: 2x = 5. Divide both sides by 2 and simplify: x = 5/2. The second solution for x is 5/2.

Step 5

Set the third term to equal zero: x + 4 = 0. Subtract 4 from both sides and simplify: x = -4, which is the third solution.