Verified answer

If x-3 is a factor then 3 is a root.

divide y by (x-3) giving a quadratic

factor and solve the quadratic

use either long division or synthetic division or try factoring by grouping.

y = x³-5x²+8x-6

but what is y(3)?

= 27-45+24-6 = 51-51 = 0 ok

3|1 -5 8 -6

_|__3 -6 6

_|1 -2 2| 0

x²-2x+2 = 0

x²-2x+1= -2+1

(x-1)² = -1

x = 1±i

x = 3,1+i,1-i

x³-5x²+8x-6

divided factor is (x-3)

x = 3

f(x) = x³-5x²+8x-6

f(3) = 27-5(9)+8(3)-6

= 27-45+24-6

f(3) = 0

the f(x) factor is (x-3)

then

x = 3 1 -5 8 -6

0 3 -6 6

_______________

1 -2 2 0

x²-2x+2 = 0

x = -b±√b²-4ac/2a

x = -(-2)±√(-2)²-4(2)/2(1)

= 2±√4-8/2

= 2±i√(2)/2

= 1±i

x³ - 5x² + 8x - 6 = 0

x³ - (2x² + 3x²) + (2x + 6x) - 6 = 0

x³ - 2x² - 3x² + 2x + 6x - 6 = 0

x³ - 2x² + 2x - 3x² + 6x - 6 = 0

(x³ - 2x² + 2x) - (3x² - 6x + 6) = 0

x.(x² - 2x + 2) - 3.(x² - 2x + 2) = 0

(x - 3).[x² - 2x + 2] = 0

(x - 3).[x² - 2x + 1 + 1] = 0

(x - 3).[(x² - 2x + 1) + 1] = 0

(x - 3).[(x - 1)² + 1] = 0 → you know that: (x - 1)² ≥ 0 because it's a square, so: [(x - 1)² + 1] ≠ 0

(x - 3) = 0

x = 3 ← this is the unique root

But, if you want to use the complex number:

(x - 3).[(x - 1)² + 1] = 0

First case: (x - 3) = 0 → x = 3

Second case: [(x - 1)² + 1] = 0

(x - 1)² + 1 = 0

(x - 1)² = - 1

(x - 1)² = i²

(x - 1)² = (± i)²

x - 1 = ± i

x = 1 ± i

→ Solution = { 3 ; (1 + i) ; (1 - i) }

Use synthetic division to get the quotient, and apply the quadratic formula to find the two complex roots.

Using long division or synthetic division, you find the other factor to be x^2 -2x +2. That has complex roots 1±i (by your favorite quadratic solution method).

The roots are x = {3, 1+i, 1-i}.

Since (x - 3) is a factor, we can divide the polynomial by (x - 3). Doing so, we end up with the quotient

x² - 2x + 2

Note that the above quadratic has complex roots. You can use the quadratic formula to find the complex roots. I ll let you take it from here.

By synthetic division:

(x^3 - 5x^2 + 8x - 6)/(x - 3) => as follows:

..... 1 ....... -5 ........ 8 ....... -6

..3.| ↓ ....... 3 ........ -6 ........ 6

...... 1 ...... -2 ........ 2 ......... 0 => Remainder

= x^2 - 2x + 2

hence:

(x^3 - 5x^2 + 8x - 6)=(x - 3) (x^2 - 2x + 2)

x = 3 => the only real root

x^2 - 2x + 2 = 0

x^2 - 2x + 1 = -2 + 1

(x - 1)^2 = -1

x - 1 = ± i

x = 1 ± i => complex roots.

x^3 – 5x^2 + 8x – 6

= x^3 - 3x^2 - 2x^2 + 6x + 2x – 6

= x^2(x-3) - 2x(x-3) + 2(x-3)

= (x-3)(x^2 - 2x + 2)

The roots of the quadratic factor are ( 2 +- sqrt((-2)^2 - 4*1*2))/2

= ( 2+- sqrt(-4))/2

= 1+-i