Solving a Linear Polynomial
Determine whether you have a linear polynomial. A linear polynomial is a polynomial of the first degree. This means that no variable will have an exponent greater than one. Because this is a first-degree polynomial, it will have exactly one real root, or solution. For example, 5 x + 2 {\displaystyle 5x+2} 5x+2 is a linear polynomial, because the variable x {\displaystyle x} x has no exponent (which is the same as an exponent of 1).
Set the polynomial equal to zero. This is a necessary step in solving all polynomials. For example, 5 x + 2 = 0 {\displaystyle 5x+2=0} 5x+2=0
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. For example, to isolate the x {\displaystyle x} x term in 5 x + 2 = 0 {\displaystyle 5x+2=0} 5x+2=0, you would subtract 2 {\displaystyle 2} 2 from both sides of the equation: 5 x + 2 = 0 {\displaystyle 5x+2=0} 5x+2=0 5 x + 2 − 2 = 0 − 2 {\displaystyle 5x+2-2=0-2} 5x+2-2=0-2 5 x = − 2 {\displaystyle 5x=-2} 5x=-2
Solve for the variable. Usually you will need to divide each side of the equation by the coefficient. This will give you the root, or solution, to your polynomial. For example, to solve for x {\displaystyle x} x in 5 x = − 2 {\displaystyle 5x=-2} 5x=-2, you would divide each side of the equation by 5 {\displaystyle 5} 5: 5 x = − 2 {\displaystyle 5x=-2} 5x=-2 5 x 5 = − 2 5 {\displaystyle {\frac {5x}{5}}={\frac {-2}{5}}} {\frac {5x}{5}}={\frac {-2}{5}} x = − 2 5 {\displaystyle x={\frac {-2}{5}}} x={\frac {-2}{5}}So, the solution to 5 x + 2 {\displaystyle 5x+2} 5x+2 is x = − 2 5 {\displaystyle x={\frac {-2}{5}}} x={\frac {-2}{5}}.
Solving a Quadratic Polynomial
Determine whether you have a quadratic polynomial. A quadratic polynomial is a polynomial of the second degree. This means that no variable will have an exponent greater than 2. Because this is a second-degree polynomial, it will have two real roots, or solutions. For example, x 2 + 8 x − 20 {\displaystyle x^{2}+8x-20} x^{{2}}+8x-20 is a quadratic polynomial, because the variable x {\displaystyle x} x has an exponent of 2 {\displaystyle 2} 2.
Make sure the polynomial is written in order of degree. This means that the term with the exponent of 2 {\displaystyle 2} 2 is listed first, followed by the first-degree term, followed by the constant. For example, you would rewrite 8 x + x 2 − 20 {\displaystyle 8x+x^{2}-20} 8x+x^{{2}}-20 as x 2 + 8 x − 20 {\displaystyle x^{2}+8x-20} x^{{2}}+8x-20.
Set the equation to equal zero. This is a necessary step for solving all polynomials. For example, x 2 + 8 x − 20 = 0 {\displaystyle x^{2}+8x-20=0} x^{{2}}+8x-20=0.
Rewrite the expression as a four-term expression. To do this, split up the first-degree term (the x {\displaystyle x} x term). You are looking for two numbers whose sum is equal to the first degree coefficient, and whose product is equal to the constant. For example, for the quadratic polynomial x 2 + 8 x − 20 = 0 {\displaystyle x^{2}+8x-20=0} x^{{2}}+8x-20=0, you need to find two numbers ( a {\displaystyle a} a and b {\displaystyle b} b), where a + b = 8 {\displaystyle a+b=8} a+b=8, and a ⋅ b = − 20 {\displaystyle a\cdot b=-20} a\cdot b=-20. Since you have − 20 {\displaystyle -20} -20, you know that one of the number will be negative. You should see that 10 + ( − 2 ) = 8 {\displaystyle 10+(-2)=8} 10+(-2)=8 and 10 ⋅ ( − 2 ) = − 20 {\displaystyle 10\cdot (-2)=-20} 10\cdot (-2)=-20. Thus, you will split up 8 x {\displaystyle 8x} 8x into 10 x − 2 x {\displaystyle 10x-2x} 10x-2x and rewrite the quadratic polynomial: x 2 + 10 x − 2 x − 20 = 0 {\displaystyle x^{2}+10x-2x-20=0} x^{{2}}+10x-2x-20=0.
Factor by grouping. To do this, factor out a term common to the first two terms in the polynomial. For example, the first two terms in the polynomial x 2 + 10 x − 2 x − 20 = 0 {\displaystyle x^{2}+10x-2x-20=0} x^{{2}}+10x-2x-20=0 are x 2 + 10 x {\displaystyle x^{2}+10x} x^{{2}}+10x. A term common to both is x {\displaystyle x} x. Thus, the factored group is x ( x + 10 ) {\displaystyle x(x+10)} x(x+10).
Factor the second group. To do this, factor out a term common to the second two terms in the polynomial. For example, the second two terms in the polynomial x 2 + 10 x − 2 x − 20 = 0 {\displaystyle x^{2}+10x-2x-20=0} x^{{2}}+10x-2x-20=0 are − 2 x − 20 {\displaystyle -2x-20} -2x-20. A term common to both is − 2 {\displaystyle -2} -2. Thus, the factored group is − 2 ( x + 10 ) {\displaystyle -2(x+10)} -2(x+10).
Rewrite the polynomial as two binomials. A binomial is a two-term expression. You already have one binomial, which is the expression in parentheses for each group. This expression should be the same for each group. The second binomial is created by combining the two terms that were factored out of each group. For example, after factoring by grouping, x 2 + 10 x − 2 x − 20 = 0 {\displaystyle x^{2}+10x-2x-20=0} x^{{2}}+10x-2x-20=0 becomes x ( x + 10 ) − 2 ( x + 10 ) = 0 {\displaystyle x(x+10)-2(x+10)=0} x(x+10)-2(x+10)=0. The first binomial is ( x + 10 ) {\displaystyle (x+10)} (x+10). The second binomial is ( x − 2 ) {\displaystyle (x-2)} (x-2). So the original quadratic polynomial, x 2 + 8 x − 20 = 0 {\displaystyle x^{2}+8x-20=0} x^{{2}}+8x-20=0 can be written as the factored expression ( x + 10 ) ( x − 2 ) = 0 {\displaystyle (x+10)(x-2)=0} (x+10)(x-2)=0.
Find the first root, or solution. To do this, solve for x {\displaystyle x} x in the first binomial. For example, to find the first root for ( x + 10 ) ( x − 2 ) = 0 {\displaystyle (x+10)(x-2)=0} (x+10)(x-2)=0, you would first set the first binomial expression to 0 {\displaystyle 0} {\displaystyle 0} and solve for x {\displaystyle x} x. Thus: x + 10 = 0 {\displaystyle x+10=0} x+10=0 x + 10 − 10 = 0 − 10 {\displaystyle x+10-10=0-10} x+10-10=0-10 x = − 10 {\displaystyle x=-10} x=-10So, the first root of the quadratic polynomial x 2 + 8 x − 20 = 0 {\displaystyle x^{2}+8x-20=0} x^{{2}}+8x-20=0 is − 10 {\displaystyle -10} -10.
Find the second root, or solution. To do this, solve for x {\displaystyle x} x in the second binomial. For example, to find the second root for ( x + 10 ) ( x − 2 ) = 0 {\displaystyle (x+10)(x-2)=0} (x+10)(x-2)=0, you would set the second binomial expression to 0 {\displaystyle 0} {\displaystyle 0} and solve for x {\displaystyle x} x. Thus: x − 2 = 0 {\displaystyle x-2=0} x-2=0 x − 2 + 2 = 0 + 2 {\displaystyle x-2+2=0+2} x-2+2=0+2 x = 2 {\displaystyle x=2} x=2So, the second root of the quadratic polynomial x 2 + 8 x − 20 = 0 {\displaystyle x^{2}+8x-20=0} x^{{2}}+8x-20=0 is 2 {\displaystyle 2} 2.
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