For example, 5x+2{\displaystyle 5x+2} is a linear polynomial, because the variable x{\displaystyle x} has no exponent (which is the same as an exponent of 1).
For example, 5x+2=0{\displaystyle 5x+2=0}
For example, to isolate the x{\displaystyle x} term in 5x+2=0{\displaystyle 5x+2=0}, you would subtract 2{\displaystyle 2} from both sides of the equation:5x+2=0{\displaystyle 5x+2=0}5x+2−2=0−2{\displaystyle 5x+2-2=0-2}5x=−2{\displaystyle 5x=-2}
For example, to solve for x{\displaystyle x} in 5x=−2{\displaystyle 5x=-2}, you would divide each side of the equation by 5{\displaystyle 5}:5x=−2{\displaystyle 5x=-2}5x5=−25{\displaystyle {\frac {5x}{5}}={\frac {-2}{5}}}x=−25{\displaystyle x={\frac {-2}{5}}}So, the solution to 5x+2{\displaystyle 5x+2} is x=−25{\displaystyle x={\frac {-2}{5}}}.
For example, x2+8x−20{\displaystyle x^{2}+8x-20} is a quadratic polynomial, because the variable x{\displaystyle x} has an exponent of 2{\displaystyle 2}.
For example, you would rewrite 8x+x2−20{\displaystyle 8x+x^{2}-20} as x2+8x−20{\displaystyle x^{2}+8x-20}.
For example, x2+8x−20=0{\displaystyle x^{2}+8x-20=0}.
For example, for the quadratic polynomial x2+8x−20=0{\displaystyle x^{2}+8x-20=0}, you need to find two numbers (a{\displaystyle a} and b{\displaystyle b}), where a+b=8{\displaystyle a+b=8}, and a⋅b=−20{\displaystyle a\cdot b=-20}. Since you have −20{\displaystyle -20}, you know that one of the number will be negative. You should see that 10+(−2)=8{\displaystyle 10+(-2)=8} and 10⋅(−2)=−20{\displaystyle 10\cdot (-2)=-20}. Thus, you will split up 8x{\displaystyle 8x} into 10x−2x{\displaystyle 10x-2x} and rewrite the quadratic polynomial: x2+10x−2x−20=0{\displaystyle x^{2}+10x-2x-20=0}.
For example, the first two terms in the polynomial x2+10x−2x−20=0{\displaystyle x^{2}+10x-2x-20=0} are x2+10x{\displaystyle x^{2}+10x}. A term common to both is x{\displaystyle x}. Thus, the factored group is x(x+10){\displaystyle x(x+10)}.
For example, the second two terms in the polynomial x2+10x−2x−20=0{\displaystyle x^{2}+10x-2x-20=0} are −2x−20{\displaystyle -2x-20}. A term common to both is −2{\displaystyle -2}. Thus, the factored group is −2(x+10){\displaystyle -2(x+10)}.
For example, after factoring by grouping, x2+10x−2x−20=0{\displaystyle x^{2}+10x-2x-20=0} becomes x(x+10)−2(x+10)=0{\displaystyle x(x+10)-2(x+10)=0}. The first binomial is (x+10){\displaystyle (x+10)}. The second binomial is (x−2){\displaystyle (x-2)}. So the original quadratic polynomial, x2+8x−20=0{\displaystyle x^{2}+8x-20=0} can be written as the factored expression (x+10)(x−2)=0{\displaystyle (x+10)(x-2)=0}.
For example, to find the first root for (x+10)(x−2)=0{\displaystyle (x+10)(x-2)=0}, you would first set the first binomial expression to 0{\displaystyle 0} and solve for x{\displaystyle x}. Thus:x+10=0{\displaystyle x+10=0}x+10−10=0−10{\displaystyle x+10-10=0-10}x=−10{\displaystyle x=-10}So, the first root of the quadratic polynomial x2+8x−20=0{\displaystyle x^{2}+8x-20=0} is −10{\displaystyle -10}.
For example, to find the second root for (x+10)(x−2)=0{\displaystyle (x+10)(x-2)=0}, you would set the second binomial expression to 0{\displaystyle 0} and solve for x{\displaystyle x}. Thus:x−2=0{\displaystyle x-2=0}x−2+2=0+2{\displaystyle x-2+2=0+2}x=2{\displaystyle x=2}So, the second root of the quadratic polynomial x2+8x−20=0{\displaystyle x^{2}+8x-20=0} is 2{\displaystyle 2}.
