Answer :
Given function, f(x) = x^6 - 12x^5 + 45x^4 - 405x^2 + 972x - 729; 3 (mult. 5).
Zeroes of f(x) are the values of x for which f(x) = 0. So, f(x) is factorable if and only if we can find zeroes of f(x).
Let's solve f(x) = 0 using x = 3 as the initial guess. Then, f(3) = 3^6 - 12(3^5) + 45(3^4) - 405(3^2) + 972(3) - 729 = 0. So, x = 3 is a zero of f(x) of the given multiplicity, which is 5.
Since x = 3 is a zero of f(x) of multiplicity 5, we can represent f(x) as follows:
$$f(x) = (x-3)^5 p(x)$$
where p(x) = EXPRESSF[X] and EXPRESSF[X] is a polynomial expression in x.
Now, we have to find the polynomial expression p(x) so that we can express f(x) as a product of linear factors.
The best way to find p(x) is by polynomial division:
$$\begin{array}{r|rrrrrr} &x^5&-5x^4&30x^3&-90x^2&180x&-243\\hline x-3&x^6&-12x^5&45x^4&-405x^2&972x&-729\\hline &x^6&-3x^5&+18x^4&-45x^3&135x^2&-243x\ & & & &360x^3&-1080x^2&648x\ & & & &360x^3&-1080x^2&648x\ & & & & &1260x^2&-891x\ & & & & &1260x^2&-3780x\ & & & & & &2889x\\end{array}$$
So, p(x) = x^5 - 3x^4 + 18x^3 - 45x^2 + 135x - 243.
Therefore, we can express f(x) as a product of linear factors as follows:
$$\begin{aligned}f(x) &= (x-3)^5 p(x)\ &= (x-3)^5 (x^5 - 3x^4 + 18x^3 - 45x^2 + 135x - 243)\ &= (x-3)^5 (x-3) (x^4 + 2x^3 + 12x^2 + 36x + 81)\ &= (x-3)^6 (x^4 + 2x^3 + 12x^2 + 36x + 81)\ \end{aligned}$$
Therefore, f(x) is a product of linear factors.
For further information on polynomial expressions, refer below:
https://brainly.com/question/29714265
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