Polynomial Equation

Polynomial Equation

Polynomial Equation

Let f(x) = anxn + an-1xn-1 + … + a0 be a polynomial in x. Then f(x)=0 is called a polynomial equation in x.

ax+b = 0 is a linear equation

ax2+bx+c = 0 is a quadratic equation

ax3+ bx2+cx+d = 0 is a cubic equation

ax4+bx3+cx2+dx+e = 0 is a big biquadratic equation

(Here, a, b, c, d, e are real numbers.)

If α is a real number such that f(α) = 0, then α is called a root of the polynomial f(x) = 0.

For example:

Consider a quadratic equation 3x2 + 5x – 2 = 0

When x = -2, then

           3x2 + 5x – 2 = 3(-2)2 + 5(-2) – 2 = 12 – 10 – 2 = 0

So, x = -2 is a root of polynomial equation 3x2 + 5x – 2 = 0


Workout Examples

Example 1: Solve: 2x3 + 13x2 – 36 = 0

Solution: Here,

                        2x3 + 13x2 – 36 = 0

            or,        2x3 + 4x2 + 9x2 + 18x – 18x – 36 = 0

            or,        2x2(x + 2) + 9x(x + 2) - 18(x + 2) = 0

            or,        (x + 2)(2x2 + 9x – 18) = 0

            or,        (x + 2)(2x2 + 12x – 3x – 18) = 0

            or,        (x + 2){2x(x +6) – 3(x + 6)} = 0

            or,        (x + 2)(x + 6)(2x – 3) = 0

            Either,           x + 2 = 0 ……………….. (i)

             Or,      x + 6 = 0 ……………….. (ii)

             Or,      2x – 3 = 0 ………………. (iii)

                        From (i), x = -2

                        From (ii), x = -6

                        From (iii), x = 3/2

Hence, x = -2, -6, 3/2

 

Example 2: Solve: x3 - 4x2 + 5x – 2 = 0

Solution: Let,

                        f(x) = x3 - 4x2 + 5x – 2

                        When x = 1, f(x) = 1 – 4 + 5 – 2 = 0

                        (x–1) is a factor of f(x).

                        Let us use synthetic division to get other factor.                       

synthetic division

                        Quotient is x2 – 3x + 2

                        Now,

                                    x2 – 3x + 2

                                    = x2 – 2x – x + 2

                                    = x(x – 2) – 1(x – 2)

                                    = (x – 2)(x – 1)

                        f(x) = x3 – 4x2 + 5x – 2

                                    = (x - 1)(x – 1)(x – 2)

                        Now,

                                    (x – 1)(x – 1)(x – 2) = 0

                        Either,           x – 1 = 0 ………………….. (i)

                        Or,       x – 1 = 0 ………………….. (ii)

                        Or,       x – 2 = 0 ………………….. (iii)

                                    From (i), x = 1

                                    From (ii), x = 1

                                    From (iii), x = 2

                        Hence, x = 1, 1, 2

           

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