Factorize the following polynomial expression: 2x^(6)-8x^(5)+12x^(4)-16x^(3)+24x^(2)-32 x

Pull out common factor: First, let's pull out the common factor from all terms. 2x^(6)-8x^(5)+12x^(4)-16x^(3)+24x^(2)-32x = 2x(x^(5)-4x^(4)+6x^(3)-8x^(2)+12x-16).Factor by grouping: Now, notice that the coefficients form a pattern, they are all multiples of 4, and the exponents decrease by 1 each time. Let's factor by grouping. 2x[(x^(5)-4x^(4))+(6x^(3)-8x^(2))+(12x-16)].Factor out common terms: Factor out the common x^4 from the first group, x^2 from the second group, and 4 from the third group. 2x[x^4(x-4)+x^2(6x-8)+4(3x-4)].Identify common factor: We can see that (x-4) is a common factor in each group. 2x[(x^4+6x^2+4)(x-4)].Factor quadratic part: Now, let's factor the quadratic part which is a perfect square. 2x[(x^2+2)^2(x-4)].Final factorized form: So the fully factorized form of the polynomial is: 2x(x^2+2)^2(x-4).

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Q. Factorize the following polynomial expression:

\newline

2x^{6}-8x^{5}+12x^{4}-16x^{3}+24x^{2}-32x

Pull out common factor: First, let's pull out the common factor from all terms. $\newline$ $2x^{6}-8x^{5}+12x^{4}-16x^{3}+24x^{2}-32x = 2x(x^{5}-4x^{4}+6x^{3}-8x^{2}+12x-16)$ .
Factor by grouping: Now, notice that the coefficients form a pattern, they are all multiples of $4$ , and the exponents decrease by $1$ each time. $\newline$ Let's factor by grouping. $\newline$ $2x[(x^{5}-4x^{4})+(6x^{3}-8x^{2})+(12x-16)]$ .
Factor out common terms: Factor out the common $x^4$ from the first group, $x^2$ from the second group, and $4$ from the third group. $\newline$ $2x[x^4(x-4)+x^2(6x-8)+4(3x-4)]$ .
Identify common factor: We can see that $(x-4)$ is a common factor in each group. $2x[(x^4+6x^2+4)(x-4)]$ .
Factor quadratic part: Now, let's factor the quadratic part which is a perfect square. $2x[(x^2+2)^2(x-4)]$ .
Final factorized form: So the fully factorized form of the polynomial is: $2x(x^2+2)^2(x-4)$ .