Answer :
To factorize the given expressions, we will proceed step-by-step for each part.
(i) Factorize [tex]8x^3 + 12x^2 + 6x + 1[/tex]
We will attempt to factorize by grouping:
Group the terms: [tex](8x^3 + 12x^2) + (6x + 1)[/tex]
Factor out the greatest common factor (GCF) from each group:
- From [tex]8x^3 + 12x^2[/tex], factor out [tex]4x^2[/tex]: [tex]4x^2(2x + 3)[/tex]
- From [tex]6x + 1[/tex], there is no common factor. We keep it as it is: [tex]6x + 1[/tex].
Now, it looks like this: [tex]4x^2(2x + 3) + 1(6x + 1)[/tex]
Notice that we cannot find a common binomial factor between the groups. Therefore, this trinomial does not factor neatly over the real numbers. In such a case, you could consider finding roots or using numerical methods for approximation if you're looking for further simplification.
(iii) Factorize [tex]x^3 + 48x^2y + 108xy^2 + 216y^3[/tex]
This expression can be approached as a sum of cubes.
Recognize that each term contains a power related to the cubes ([tex]x^3[/tex], [tex]y^3[/tex]). Notice the structure resembles a polynomial generated from binomials.
Look for factorization patterns or examine if trying a grouping strategy or the use of synthetic division might reveal simple linear or quadratic factors. Upon recognition, observe:
- The polynomial can be rewritten as: [tex](x + 6y)^3[/tex].
This occurs because:
- Expanding [tex](x + 6y)^3[/tex] gives [tex]x^3 + 3(x^2)(6y) + 3(x)(36y^2) + (216y^3)[/tex], which correspond to the terms in the polynomial. Therefore, we can factorize using the simplification as:
[tex]x^3 + 48x^2y + 108xy^2 + 216y^3 = (x + 6y)^3[/tex]
By recognizing these patterns, we've factorized the second expression efficiently.
