Question

activity 9 finding the roots of polynomial equations
find all real roots of the following equations. Next, write each polynomial on the left side of the equation in factored form. show your complete solutions

1. x^3-10x^2+32x-32=0
2. x^3-6x^2+11x-6=0
3.x^3-2x^2+4x-8=0
4.3x^3-19x^2+33x-9=0
5. x^4-5x^2+4=0

Answer 1

Answer:

x³ -10x² + 32x - 32 = (x - 4)²(x-2)

x³ -6x² + 11x - 6 = (x - 1)(x-2)(x-3)

x³ - 2x² + 4x - 8 =  (x - 2)(x² + 4)

3x³ - 19x² + 33x - 9 =  (3x - 1)(x - 3)²

x⁴ - 5x² + 4 =  (x + 1)(x - 1) (x + 2)(x - 2)

Step-by-step explanation:

x³ -10x² + 32x - 32 = 0

=> x³ - 2x² - 8x² + 16x + 16x - 32 = 0

=> x²(x - 2) - 8x(x - 2) + 16(x - 2) = 0

=> (x² - 8x + 16)(x-2) = 0

=> (x - 4)²(x-2) = 0

x = 2 , 4 , 4

x³ -10x² + 32x - 32 = (x - 4)²(x-2)

x³ -6x² + 11x - 6 = 0

=> x³ -x²  -5x² + 5x + 6x - 6 = 0

=> x²(x - 1) - 5x(x - 1) + 6(x - 1) = 0

=> (x - 1)(x² - 5x + 6) = 0

=> (x - 1) (x² - 2x - 3x + 6) = 0

=> (x - 1)(x-2)(x-3) = 0

x = 1 , 2 , 3

x³ -6x² + 11x - 6 = (x - 1)(x-2)(x-3)

x³ - 2x² + 4x - 8 = 0

=>x²(x - 2) + 4(x - 2) = 0

=> (x - 2)(x² + 4) = 0

x = 2

x³ - 2x² + 4x - 8 =  (x - 2)(x² + 4)

3x³ - 19x² + 33x - 9 = 0

=> 3x³ - x² - 18x² + 6x + 27x - 9 = 0

=> x²(3x - 1) - 6x(3x - 1) + 9(3x - 1) = 0

=> (3x-1)(x² - 6x + 9) = 0

=> (3x - 1)(x - 3)² = 0

x = 1/3 , 3

3x³ - 19x² + 33x - 9 =  (3x - 1)(x - 3)²

x⁴ - 5x² + 4 = 0

=> x⁴ - x² - 4x² + 4 = 0

=> x²(x² - 1) - 4(x²- 1) = 0

=> (x² - 1)(x² - 4) = 0

=> (x + 1)(x - 1) (x + 2)(x - 2) = 0

x = 1 , 2 , - 1 , -1

x⁴ - 5x² + 4 =  (x + 1)(x - 1) (x + 2)(x - 2)