Verified answer

Let f(x) = x^3−14x^2+53x−40

Since f(5) = 0, (x-5) is a factor. Using synthetic division,

5 | 1 -14 53 -40

..........5 -45 40

--------------------

.....1 -9 8 | 0

x^3−14x^2+53x−40 = (x-5)(x^2-9x+8) = (x-5)(x-8)(x-1) = 0

Answer: x = 1, 5, 8

Solution

x= 1, 5, 8

Given

x= 5 == > x-5 is a factor

factor out (x-5)

x^3 -14x^2 + 53x -40

x^2 <===

x^3-5x^2 { x^2(x-5)}

-9x^2 + 53x -40

-9x <===

-9x^2+45x { 9x(x-5)}

8x -40

8 <===

8x-40 { 8(x-5)}

0___

x^2 - 9x + 8

x^3 -14x^2 + 53x -40 =(x-5)(x^2-9x+8)

=(x-5)(x-8)(x-1)

Use long division to divide your expression by (x-5) to get a quadratic factor, so we have:

=(x² - 9x + 8)(x-5)

=(x-1)(x-8)(x-5)

Real roots, x=1, x=8, x=5

Use synthetic division.

5 | 1 . -14 . 53 . -40

---------> 5 .-45 .. 40

--> 1 .. -9 .. 8 .. . 0

(x - 5)(x² - 9x + 8) = 0

(x - 5)(x - 8)(x - 1) = 0

x = 5, 8, 1