Difference between revisions of "Solve Polynomials"
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=== Solving a Linear Polynomial === | === Solving a Linear Polynomial === | ||
| − | #Determine whether you have a linear polynomial. A linear polynomial is a polynomial to the first degree.<ref>http://www.mathwords.com/l/linear_polynomial.htm</ref> This means that no variable will have an exponent (or an exponent greater than 1). Because this is a first-degree polynomial, it will have exactly one root, or solution.<ref>https://www.math.utah.edu/~wortman/1050-text-calp.pdf</ref> | + | #Determine whether you have a linear polynomial. A linear polynomial is a polynomial to the first degree.<ref name="rf1">http://www.mathwords.com/l/linear_polynomial.htm</ref> This means that no variable will have an exponent (or an exponent greater than 1). Because this is a first-degree polynomial, it will have exactly one root, or solution.<ref name="rf2">https://www.math.utah.edu/~wortman/1050-text-calp.pdf</ref> |
#*For example, <math>5x + 2</math> is a linear polynomial, because the variable <math>x</math> has no exponent (which is the same as an exponent of 1). | #*For example, <math>5x + 2</math> is a linear polynomial, because the variable <math>x</math> has no exponent (which is the same as an exponent of 1). | ||
#Set the equation to equal zero. This is a necessary step for solving all polynomials. | #Set the equation to equal zero. This is a necessary step for solving all polynomials. | ||
#*For example, <math>5x + 2 = 0</math> | #*For example, <math>5x + 2 = 0</math> | ||
| − | #Isolate the variable term. To do this, add or subtract the constant from both sides of the equation. A constant is a term without a variable.<ref>http://www.mathwords.com/c/constant.htm</ref> | + | #Isolate the variable term. To do this, add or subtract the constant from both sides of the equation. A constant is a term without a variable.<ref name="rf3">http://www.mathwords.com/c/constant.htm</ref> |
#*For example, to isolate the <math>x</math> term in <math>5x + 2 = 0</math>, you would subtract <math> 2</math> from both sides of the equation:<br><math>5x + 2 = 0</math><br><math>5x + 2 - 2 = 0 - 2</math><br><math>5x = - 2</math> | #*For example, to isolate the <math>x</math> term in <math>5x + 2 = 0</math>, you would subtract <math> 2</math> from both sides of the equation:<br><math>5x + 2 = 0</math><br><math>5x + 2 - 2 = 0 - 2</math><br><math>5x = - 2</math> | ||
#Solve for the variable. Usually you will need to divide each side of the equation by the constant. This will give you the root, or solution, to your polynomial. | #Solve for the variable. Usually you will need to divide each side of the equation by the constant. This will give you the root, or solution, to your polynomial. | ||
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=== Solving a Quadratic Polynomial === | === Solving a Quadratic Polynomial === | ||
| − | #Determine whether you have a quadratic polynomial. A quadratic polynomial is a polynomial to the second degree.<ref>http://www.mathwords.com/q/quadratic_polynomial.htm</ref> This means that no variable will have an exponent greater than 2. Because this is a second-degree polynomial, it will have two roots, or solutions.<ref>http://www.themathpage.com/aprecalc/quadratic-equation.htm#double</ref> | + | #Determine whether you have a quadratic polynomial. A quadratic polynomial is a polynomial to the second degree.<ref name="rf4">http://www.mathwords.com/q/quadratic_polynomial.htm</ref> This means that no variable will have an exponent greater than 2. Because this is a second-degree polynomial, it will have two roots, or solutions.<ref name="rf5">http://www.themathpage.com/aprecalc/quadratic-equation.htm#double</ref> |
#*For example, <math>x^{2} + 8x - 20</math> is a quadratic polynomial, because the variable <math>x</math> has an exponent of <math> 2</math>. | #*For example, <math>x^{2} + 8x - 20</math> is a quadratic polynomial, because the variable <math>x</math> has an exponent of <math> 2</math>. | ||
| − | #Make sure the polynomial is written in order of degree. This means that the term with the exponent of <math> 2</math> is listed first, followed by the first-degree term, followed by the constant.<ref>http://www.mesacc.edu/~scotz47781/mat120/notes/factoring/trinomials/a_is_not_1/trinomials_a_is_not_1.html</ref> | + | #Make sure the polynomial is written in order of degree. This means that the term with the exponent of <math> 2</math> is listed first, followed by the first-degree term, followed by the constant.<ref name="rf6">http://www.mesacc.edu/~scotz47781/mat120/notes/factoring/trinomials/a_is_not_1/trinomials_a_is_not_1.html</ref> |
#*For example, you would rewrite <math>8x +x^{2} - 20</math> as <math>x^{2} + 8x - 20</math>. | #*For example, you would rewrite <math>8x +x^{2} - 20</math> as <math>x^{2} + 8x - 20</math>. | ||
#Set the equation to equal zero. This is a necessary step for solving all polynomials. | #Set the equation to equal zero. This is a necessary step for solving all polynomials. | ||
#*For example, <math>x^{2} + 8x - 20 = 0</math>. | #*For example, <math>x^{2} + 8x - 20 = 0</math>. | ||
| − | #Rewrite the expression as a four-term expression. To do this, split up the first-degree term (the <math>x</math> term). You are looking for two numbers whose sum is equal to the first degree coefficient, and whose product is equal to the constant. <ref>https://www.khanacademy.org/math/algebra/quadratics/solving-quadratic-equations-by-factoring/v/example-1-solving-a-quadratic-equation-by-factoring</ref> | + | #Rewrite the expression as a four-term expression. To do this, split up the first-degree term (the <math>x</math> term). You are looking for two numbers whose sum is equal to the first degree coefficient, and whose product is equal to the constant. <ref name="rf7">https://www.khanacademy.org/math/algebra/quadratics/solving-quadratic-equations-by-factoring/v/example-1-solving-a-quadratic-equation-by-factoring</ref> |
#*For example, for the quadratic polynomial <math>x^{2} + 8x - 20 = 0</math>, you need to find two numbers (<math>a</math> and <math>b</math>), where <math>a + b = 8</math>, and <math>a \cdot b = -20</math>. | #*For example, for the quadratic polynomial <math>x^{2} + 8x - 20 = 0</math>, you need to find two numbers (<math>a</math> and <math>b</math>), where <math>a + b = 8</math>, and <math>a \cdot b = -20</math>. | ||
#*Since you have <math>-20</math>, you know that one of the number will be negative. | #*Since you have <math>-20</math>, you know that one of the number will be negative. | ||
#*You should see that <math>10 + (-2) = 8</math> and <math>10 \cdot (-2) = -20</math>. Thus, you will split up <math>8x</math> into <math>10x - 2x</math> and rewrite the quadratic polynomial: <math>x^{2} + 10x - 2x - 20 = 0</math>. | #*You should see that <math>10 + (-2) = 8</math> and <math>10 \cdot (-2) = -20</math>. Thus, you will split up <math>8x</math> into <math>10x - 2x</math> and rewrite the quadratic polynomial: <math>x^{2} + 10x - 2x - 20 = 0</math>. | ||
| − | #Factor by grouping. To do this, factor out a term common to the first two terms in the polynomial.<ref | + | #Factor by grouping. To do this, factor out a term common to the first two terms in the polynomial.<ref name="rf6" /> |
#*For example, the first two terms in the polynomial <math>x^{2} + 10x - 2x - 20 = 0</math> are <math>x^{2} + 10x</math>. A term common to both is <math>x</math>. Thus, the factored group is <math>x(x + 10)</math>. | #*For example, the first two terms in the polynomial <math>x^{2} + 10x - 2x - 20 = 0</math> are <math>x^{2} + 10x</math>. A term common to both is <math>x</math>. Thus, the factored group is <math>x(x + 10)</math>. | ||
#Factor the second group. To do this, factor out a term common to the second two terms in the polynomial. | #Factor the second group. To do this, factor out a term common to the second two terms in the polynomial. | ||
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#*The second binomial is <math>(x - 2)</math>. | #*The second binomial is <math>(x - 2)</math>. | ||
#*So the original quadratic polynomial, <math>x^{2} + 8x - 20 = 0</math> can be written as the factored expression <math>(x + 10)(x - 2) = 0</math>. | #*So the original quadratic polynomial, <math>x^{2} + 8x - 20 = 0</math> can be written as the factored expression <math>(x + 10)(x - 2) = 0</math>. | ||
| − | #Find the first root, or solution. To do this, solve for <math>x</math> in the first binomial.<ref | + | #Find the first root, or solution. To do this, solve for <math>x</math> in the first binomial.<ref name="rf7" /> |
#*For example, to find the first root for <math>(x + 10)(x - 2) = 0</math>, you would first set the first binomial expression to <math>0</math> and solve for <math>x</math>. Thus:<br><math>x + 10 = 0</math><br><math>x + 10 - 10 = 0 - 10</math><br><math>x = - 10</math><br>So, the first root of the quadratic polynomial <math>x^{2} + 8x - 20 = 0</math> is <math>- 10</math>. | #*For example, to find the first root for <math>(x + 10)(x - 2) = 0</math>, you would first set the first binomial expression to <math>0</math> and solve for <math>x</math>. Thus:<br><math>x + 10 = 0</math><br><math>x + 10 - 10 = 0 - 10</math><br><math>x = - 10</math><br>So, the first root of the quadratic polynomial <math>x^{2} + 8x - 20 = 0</math> is <math>- 10</math>. | ||
| − | #Find the second root, or solution. To do this, solve for <math>x</math> in the second binomial.<ref | + | #Find the second root, or solution. To do this, solve for <math>x</math> in the second binomial.<ref name="rf7" /> |
#*For example, to find the second root for <math>(x + 10)(x - 2) = 0</math>, you would set the second binomial expression to <math>0</math> and solve for <math>x</math>. Thus:<br><math>x - 2 = 0</math><br><math>x - 2 + 2 = 0 + 2</math><br><math>x = 2</math><br>So, the second root of the quadratic polynomial <math>x^{2} + 8x - 20 = 0</math> is <math> 2</math>. | #*For example, to find the second root for <math>(x + 10)(x - 2) = 0</math>, you would set the second binomial expression to <math>0</math> and solve for <math>x</math>. Thus:<br><math>x - 2 = 0</math><br><math>x - 2 + 2 = 0 + 2</math><br><math>x = 2</math><br>So, the second root of the quadratic polynomial <math>x^{2} + 8x - 20 = 0</math> is <math> 2</math>. | ||
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== Tips == | == Tips == | ||
*Don't fret if you get different variables, like t, or if you see an equation set to f(x) instead of 0. If the question wants roots, zeros, or factors, just treat it like any other problem. | *Don't fret if you get different variables, like t, or if you see an equation set to f(x) instead of 0. If the question wants roots, zeros, or factors, just treat it like any other problem. | ||
| − | *Remember the order of operations while you work -- First work in the parenthesis, then do the multiplication and division, and finally do the addition and subtraction.<ref>http://www.mathgoodies.com/lessons/vol7/order_operations.html</ref> | + | *Remember the order of operations while you work -- First work in the parenthesis, then do the multiplication and division, and finally do the addition and subtraction.<ref name="rf8">http://www.mathgoodies.com/lessons/vol7/order_operations.html</ref> |
== Related Articles == | == Related Articles == | ||
