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Solving polynomial equations

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Presentation on theme: "Solving polynomial equations"— Presentation transcript:

1 Solving polynomial equations
Lesson 66 Solving polynomial equations

2 Using factoring to solve polynomial equations
Solve each equation: 2x4 - 10x3 - 72x2 = 0 Factor GCF x2(x2-5x -36) = 0 Factor trinomial x2(x-9)(x+4) = 0 Zero product prop. 2x2 = 0, x = 0 x-9 = 0 , x = 9 x+4 = 0 , x = -4

3 Solve by factoring 4x3 - 4x2 + x = 0 3x3 - 5x2 - 2x = 0 x3 - 12x = x2

4 Multiplicity of roots The multiplicity of a root r of a polynomial equation is the number of times x-r is a factor of the polynomial. Find the roots of each equation: x5 + 8x4 + 16x3 = 0 x3( x2 + 8x + 16) = 0 x3 ( x+4)(x+4) = 0, x3 = 0, x= 0 because x3 = xxx, root 0 has multiplicity of 3 x+4 = 0, x= -4 x+4 = 0, x= -4 because x + 4 is a factor twice, the root -4 has a multiplicity of 2

5 Find the roots and the multiplicity of each root?
x4 = 2x2 -1 x4 - 2x2 + 1= 0 (x2-1)(x2 -1) = 0 (x-1)(x+1) (x-1)(x+1) = 0 x-1 = 0 or x+1 = 0 or x-1 =0 or x+1 = 0 x = x = x = x = -1 Roots are 1 and -1, because both occur twice, they each have a multiplicity of 2

6 Find the roots and multiplicity
x5 - 12x4 + 36x3 = 0 x4 = 8x2 - 6

7 Graph the previous equations
What happens when the root's multiplicity is odd? What happens when the root's multiplicity is even?

8 A polynomial equation like
x3 - 2x2 - 11x + 12 = 0, where the degree of the polynomial is greater than 2, can be factored by using synthetic division, if one of the factors is a linear binomial and is already known. When none of the factors are given, begin factoring by using the Rational Root Theorem

9 The rational root theorem
If a polynomial P(x) has integer coefficients, then every rational root of P(x) = 0 can be written in the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient of P(x)

10 The irrational root theorem
If a polynomial P(x) has rational coefficients, and a is a root of P(x) =0, where a and b are rational and is irrational, then a is also a root of P(x) = 0

11 Identifying all real roots of a polynomial equation
find real roots of x3 - 2x2 - 11x + 12 = 0 Find all factors of the constant term: +1, -1, +2, -2,+3,-3,+4, -4, +6, -6, +12, -12 + 1 Use synthetic division to test roots until a factor is found x-1 is a factor so (x-1)(x2 - x -12) =0 (x-1)(x-4)(x+3) = 0 x = 1, x = 4, x = -3

12 Identify all real roots
3x4 - 5x3 - 29x2 + 3x + 4 = 0 Use synthetic division to test divisors for factors until you get remainder of 0. P(4) = 0 So x-4 is a factor (x-4)(3x3 + 7x2 - x - 1) = 0 Use synthetic division to test for factors of 3x3+ 7x2 -x -1 So x + 1/3 is a factor and 3x2 + 6x -1 is the other factor Use quadratic formula to factor the quadratic

13 Identify all real roots
x3 - 2x2 - 5x + 6 = 0 4x4 - 29x3 + 39x2 + 32x - 10

14 using a calculator to solve polynomial equations
Graph Use the CALC command to find the x-intercepts If you find a rational root, use this to do synthetic division, then use the quadratic formula to find the other roots. The calculator gives approximations of the exact roots.

15 Solve with your calculator
x3 - x2 - 13x - 3 = 0 x3 + 5x2 - 9x - 10 = 0


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