Answer :
Let's resolve each expression into its factors step by step.
(a) [tex]\(1 + b^2 + b^4\)[/tex]
To factor this expression, realize that it forms a quadratic in disguise:
[tex]\[ b^4 + b^2 + 1 = (b^2)^2 + b^2 + 1 \][/tex]
We can factor it using the quadratic method, considering it as:
[tex]\[ b^4 + b^2 + 1 = (b^2 + b + 1)(b^2 - b + 1) \][/tex]
(b) [tex]\(x^4 + 3x^2 + 4\)[/tex]
This expression resembles a quadratic in terms of [tex]\(x^2\)[/tex]:
[tex]\[ (x^2)^2 + 3x^2 + 4 \][/tex]
Applying the quadratic formula for factoring, we get:
[tex]\[ x^4 + 3x^2 + 4 = (x^2 - x + 2)(x^2 + x + 2) \][/tex]
(c) [tex]\(4x^4 + 3x^2 + 1\)[/tex]
Similarly, treat it as a quadratic in terms of [tex]\(x^2\)[/tex]:
[tex]\[ (2x^2)^2 + 3x^2 + 1 \][/tex]
Factoring gives:
[tex]\[ 4x^4 + 3x^2 + 1 = (2x^2 - x + 1)(2x^2 + x + 1) \][/tex]
(d) [tex]\(a^4 - 31a^2 + 9\)[/tex]
This expression is also a quadratic in terms of [tex]\(a^2\)[/tex]:
[tex]\[ (a^2)^2 - 31a^2 + 9 \][/tex]
Factoring it, we find:
[tex]\[ a^4 - 31a^2 + 9 = (a^2 - 5a - 3)(a^2 + 5a - 3) \][/tex]
(e) [tex]\(x^8 + x^4 y^4 + y^8\)[/tex]
This expression can be viewed as:
[tex]\[ (x^4)^2 + (x^4)(y^4) + (y^4)^2 \][/tex]
Using factoring methods for symmetric polynomials, this becomes:
[tex]\[ (x^8 + x^4 y^4 + y^8) = (x^2 - xy + y^2)(x^2 + xy + y^2)(x^4 - x^2y^2 + y^4) \][/tex]
(f) [tex]\(x^8 + 9x^4 + 81\)[/tex]
This resembles a sum of cubes form:
[tex]\[ (x^4)^2 + (3x^2)^2 + 9^2 \][/tex]
Factoring it using the sum of squares technique results in:
[tex]\[ x^8 + 9x^4 + 81 = (x^2 - 3x + 3)(x^2 + 3x + 3)(x^4 + 3x^2 + 9) \][/tex]
These are the factored forms of the given expressions.
(a) [tex]\(1 + b^2 + b^4\)[/tex]
To factor this expression, realize that it forms a quadratic in disguise:
[tex]\[ b^4 + b^2 + 1 = (b^2)^2 + b^2 + 1 \][/tex]
We can factor it using the quadratic method, considering it as:
[tex]\[ b^4 + b^2 + 1 = (b^2 + b + 1)(b^2 - b + 1) \][/tex]
(b) [tex]\(x^4 + 3x^2 + 4\)[/tex]
This expression resembles a quadratic in terms of [tex]\(x^2\)[/tex]:
[tex]\[ (x^2)^2 + 3x^2 + 4 \][/tex]
Applying the quadratic formula for factoring, we get:
[tex]\[ x^4 + 3x^2 + 4 = (x^2 - x + 2)(x^2 + x + 2) \][/tex]
(c) [tex]\(4x^4 + 3x^2 + 1\)[/tex]
Similarly, treat it as a quadratic in terms of [tex]\(x^2\)[/tex]:
[tex]\[ (2x^2)^2 + 3x^2 + 1 \][/tex]
Factoring gives:
[tex]\[ 4x^4 + 3x^2 + 1 = (2x^2 - x + 1)(2x^2 + x + 1) \][/tex]
(d) [tex]\(a^4 - 31a^2 + 9\)[/tex]
This expression is also a quadratic in terms of [tex]\(a^2\)[/tex]:
[tex]\[ (a^2)^2 - 31a^2 + 9 \][/tex]
Factoring it, we find:
[tex]\[ a^4 - 31a^2 + 9 = (a^2 - 5a - 3)(a^2 + 5a - 3) \][/tex]
(e) [tex]\(x^8 + x^4 y^4 + y^8\)[/tex]
This expression can be viewed as:
[tex]\[ (x^4)^2 + (x^4)(y^4) + (y^4)^2 \][/tex]
Using factoring methods for symmetric polynomials, this becomes:
[tex]\[ (x^8 + x^4 y^4 + y^8) = (x^2 - xy + y^2)(x^2 + xy + y^2)(x^4 - x^2y^2 + y^4) \][/tex]
(f) [tex]\(x^8 + 9x^4 + 81\)[/tex]
This resembles a sum of cubes form:
[tex]\[ (x^4)^2 + (3x^2)^2 + 9^2 \][/tex]
Factoring it using the sum of squares technique results in:
[tex]\[ x^8 + 9x^4 + 81 = (x^2 - 3x + 3)(x^2 + 3x + 3)(x^4 + 3x^2 + 9) \][/tex]
These are the factored forms of the given expressions.
