What is factor (x-3)^4-5x(x-3)^2+4 ?

The solution to factor (x-3)^4-5x(x-3)^2+4 is (x-1)(x^3-16x^2+68x-85)

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factor (x-3)^4-5x(x-3)^2+4

factor

(x - 3)^{4} - 5 x (x - 3)^{2} + 4

(x - 1) (x^{3} - 16 x^{2} + 68 x - 85)

Solution steps

(x - 3)^{4} - 5 x (x - 3)^{2} + 4

(x - 3)^{4} : x^{4} - 12 x^{3} + 54 x^{2} - 108 x + 81

= x^{4} - 12 x^{3} + 54 x^{2} - 108 x + 81 - 5 x (x - 3)^{2} + 4

(x - 3)^{2} : x^{2} - 6 x + 9

= x^{4} - 12 x^{3} + 54 x^{2} - 108 x + 81 - 5 x (x^{2} - 6 x + 9) + 4

Expand

- 5 x (x^{2} - 6 x + 9) : - 5 x^{3} + 30 x^{2} - 45 x

= x^{4} - 12 x^{3} + 54 x^{2} - 108 x + 81 - 5 x^{3} + 30 x^{2} - 45 x + 4

Simplify

x^{4} - 12 x^{3} + 54 x^{2} - 108 x + 81 - 5 x^{3} + 30 x^{2} - 45 x + 4 : x^{4} - 17 x^{3} + 84 x^{2} - 153 x + 85

= x^{4} - 17 x^{3} + 84 x^{2} - 153 x + 85

Use the rational root theorem

a_{0} = 85, a_{n} = 1

The dividers of

a_{0} : 1, 5, 17, 85,

The dividers of

a_{n} : 1

Therefore, check the following rational numbers:

\pm \frac{1 , 5 , 17 , 85}{1}

\frac{1}{1}

is a root of the expression, so factor out

x - 1

= (x - 1) \frac{x ^{4} - 17 x ^{3} + 84 x ^{2} - 153 x + 85}{x - 1}

\frac{x ^{4} - 17 x ^{3} + 84 x ^{2} - 153 x + 85}{x - 1} = x^{3} - 16 x^{2} + 68 x - 85

= (x - 1) (x^{3} - 16 x^{2} + 68 x - 85)

(\frac{2}{3} m - n)^{2}

l [e^{t} cos (x)]

- 58 e^{2} + 96

x^{9} x^{2 - 4}

d i scr iminan t e (4^{2}) - 4 (- 1) (1)

What is factor (x-3)^4-5x(x-3)^2+4 ?
The solution to factor (x-3)^4-5x(x-3)^2+4 is (x-1)(x^3-16x^2+68x-85)