# A of the polynomial equation: 2x^{3}+7x^{2}+3 Solution: Given

A polynomial is defined as the sum of more than one or more algebraic terms where each term consists of several degrees of same variables and integer coefficient to that variables. x2?3×2?3, 5×4?3×2+x?45×4?3×2+x?4 are some examples of polynomials. The roots or also called as zeroes of a polynomial P(x) are the value of x for which polynomial P(x) is equal to 0. In other words, we can say that polynomial P(x) will have the same value of x if x=r i.e. the value of the root of the polynomial that will satisfy the equation P(x) = 0. These are sometimes called solving the polynomial. The degree of the polynomial is always equal to the number of roots of polynomial P(x).DefinitionIn any polynomial, the root is that the value of the variable that satisfies the polynomial. Polynomial is an expression consisting of variables and coefficients of the form: P_{n}(x)= a_{n}x^{n}+a_{n-1}x^{n-1}+…+a_{0} , where a_{n} is not equal to zero and n refers to the degree of a polynomial and a_{0}, a_{1},…. a_{n} are real coefficient. Thus, the degree of the polynomial gives the idea of the number of roots of that polynomial. The roots may be different.Example 1: Find the roots of the polynomial equation: x^{2}+4x+4 Solution: Given polynomial equation x^{2}+4x+4 By factoring the quadratic: x^{2}+4x+4 = x^{2}+2x+2x+4 = 0 x(x+2) + 2(x+2) = 0 therefore, (x+2)(x+2)=0Set each factor equal to zero: x+2 =0 or x+2 = 0So, x=-2 or x=-2 . Both the roots are same, i.e. -2.Example 2: Find the roots of the polynomial equation: 2x^{3}+7x^{2}+3 Solution: Given polynomial equation 2x^{3}+7x^{2}+3 By factoring the quadratic: 2x^{3}+7x^{2}+3 = x(2x^{2}+6x+x+3) = 0 x(2x(x + 3) + (x + 3)) = 0 therefore, x(2x + 1)(x + 3) = 0Set each factor equal to 0: x = 0,2x+1 = 0,x+3 = 0So, x = 0,x = frac{-1}{2} ,x = -3. Zeroes of polynomial are frac{-1}{2} ,-3,0.Quadratic roots of PolynomialRoots are the solution to the polynomial. The roots may be real or complex (imaginary), and they might not be distinct. A quadratic equation is ax^{2}+bx+c=0 , where a

eq 0 and x = frac{-b pm sqrt{b^{2}-4ac}}{2a}If the coefficients a, b, c are real, it follows that: if b^{2}-4ac > 0 = the roots are real and unequal, if b^{2}-4ac = 0 = the roots are real and equal, if b^{2}-4ac < 0 the roots are imaginary.Example 1: Find the roots of the quadratic polynomial equation: x^{2}-10x+26 = 0 Solution: Given quadratic polynomial equation x^{2}-10x+26 = 0 So, a = 1,b = -10 and c = 26By putting the formula as D = b^{2}-4ac = 100 – 4 * 1 * 26 = 100 – 104 = -4 < 0Therefore D < 0,so roots are complex or imaginary.Now finding the value of x, using quadratic formula = x = frac{-b pm sqrt{b^{2}-4ac}}{2a} = frac{-(-10) pm sqrt{-4}}{2*1} = frac{10pm 2sqrt{-1}}{2} = frac{10pm 2{i}}{2} = 5 pm i Therefore, the roots are 5 + i and 5 – i.Example 2: Find the roots of the quadratic polynomial equation: x^{4}-81 Solution: Given polynomial equation: x^{4}-81 = (x^{2})^{2}-9^{2} = (x^{2}+9)(x^{2}-9) = (x^{2}+9)(x^{2}-3^{2}) = (x^{2}+9)(x+3)(x-3) .So, roots are x^{2}+9=0 = x^{2}=-9 Therefore, x = sqrt{-9} = +3i,-3i (imaginary roots) and real roots are +3,-3.ExerciseFind the roots of polynomials by factoring: x^{2}+2x-15=0 x^{4}-13x^{2}=-36 x^{2}-14x+49 x^{2}-10x+25 x^{2}+2x-15 Find the roots of the quadratic polynomial equation: x^{2}-5x+6 2x^{2}+7x-4 6y^{2}-13y+6 x^{2}+2x-8 x^{2}-4x+3