What is factor 6(x^2-4x+4)^2+(x^2-4x+4)-1 ?

The solution to factor 6(x^2-4x+4)^2+(x^2-4x+4)-1 is (3x^2-12x+11)(2x^2-8x+9)

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factor 6(x^2-4x+4)^2+(x^2-4x+4)-1

factor

6 (x^{2} - 4 x + 4)^{2} + (x^{2} - 4 x + 4) - 1

(3 x^{2} - 12 x + 11) (2 x^{2} - 8 x + 9)

Solution steps

6 (x^{2} - 4 x + 4)^{2} + (x^{2} - 4 x + 4) - 1

Break the expression into groups

= (6 (x^{2} - 4 x + 4)^{2} - 2 (x^{2} - 4 x + 4)) + (3 (x^{2} - 4 x + 4) - 1)

Factor out

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

from

6 (x^{2} - 4 x + 4)^{2} - 2 (x^{2} - 4 x + 4) : 2 (x^{2} - 4 x + 4) (3 (x^{2} - 4 x + 4) - 1)

= 2 (x^{2} - 4 x + 4) (3 (x^{2} - 4 x + 4) - 1) + (3 (x^{2} - 4 x + 4) - 1)

Factor out common term

3 (x^{2} - 4 x + 4) - 1

= (3 (x^{2} - 4 x + 4) - 1) (2 (x^{2} - 4 x + 4) + 1)

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

= (3 (x - 2)^{2} - 1) (2 (x^{2} - 4 x + 4) + 1)

2 (x^{2} - 4 x + 4) = 2 (x - 2)^{2}

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

Expand

3 (x - 2)^{2} - 1 : 3 x^{2} - 12 x + 11

= (3 x^{2} - 12 x + 11) (2 (x - 2)^{2} + 1)

Expand

2 (x - 2)^{2} + 1 : 2 x^{2} - 8 x + 9

= (3 x^{2} - 12 x + 11) (2 x^{2} - 8 x + 9)

(x - 35) (x + 61) \leq 0

(4 - 3 i)^{3}

\frac{x - 8}{2} \geq \frac{x - 4}{3} + \frac{5}{6}

1 - \frac{1}{2} + \frac{1}{3}

x^{2} - 5 x + 2 = 0

What is factor 6(x^2-4x+4)^2+(x^2-4x+4)-1 ?
The solution to factor 6(x^2-4x+4)^2+(x^2-4x+4)-1 is (3x^2-12x+11)(2x^2-8x+9)