Answer :
To factor the polynomial expression [tex]\(a^8 - 28a^4 + 36\)[/tex], let's follow a systematic process to break it down into its factors.
1. Expression Analysis:
The given expression is [tex]\(a^8 - 28a^4 + 36\)[/tex]. It is a polynomial in terms of [tex]\(a^4\)[/tex].
2. Substitution:
Let's make a substitution to simplify the expression. We will let [tex]\(u = a^4\)[/tex]. Therefore, the expression [tex]\(a^8\)[/tex] can be rewritten as [tex]\(u^2\)[/tex], since [tex]\((a^4)^2 = a^8\)[/tex].
The expression becomes:
[tex]\[ u^2 - 28u + 36 \][/tex]
3. Factoring the Quadratic Polynomial:
Now, we need to factor the quadratic polynomial [tex]\(u^2 - 28u + 36\)[/tex]. We look for two numbers that multiply to [tex]\(36\)[/tex] and add up to [tex]\(-28\)[/tex].
However, the polynomial does not factor neatly into integers, so we will solve it using other means such as completing the square or using the quadratic formula.
4. Quadratic Formula:
The quadratic formula for [tex]\(ax^2 + bx + c = 0\)[/tex] is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Applying this to [tex]\(u^2 - 28u + 36\)[/tex]:
[tex]\[ u = \frac{28 \pm \sqrt{28^2 - 4(1)(36)}}{2(1)} \][/tex]
[tex]\[ u = \frac{28 \pm \sqrt{784 - 144}}{2} \][/tex]
[tex]\[ u = \frac{28 \pm \sqrt{640}}{2} \][/tex]
[tex]\[ u = \frac{28 \pm 8\sqrt{10}}{2} \][/tex]
[tex]\[ u = 14 \pm 4\sqrt{10} \][/tex]
5. Back-Substitution:
Recall that [tex]\(u = a^4\)[/tex]. So we have two solutions for [tex]\(a^4\)[/tex]:
[tex]\[ a^4 = 14 + 4\sqrt{10} \][/tex]
[tex]\[ a^4 = 14 - 4\sqrt{10} \][/tex]
6. Final Expression:
If we were to put this back into a factorized polynomial, we can express it as:
[tex]\[ (a^4 - (14 + 4\sqrt{10}))(a^4 - (14 - 4\sqrt{10})) \][/tex]
But for simplicity and cleanliness, we recognize that the given form factors elegantly into:
[tex]\[ (a^4 - 4a^2 - 6)(a^4 + 4a^2 - 6) \][/tex]
Hence, the factored form of the expression [tex]\(a^8 - 28a^4 + 36\)[/tex] is:
[tex]\[ (a^4 - 4a^2 - 6)(a^4 + 4a^2 - 6) \][/tex]
1. Expression Analysis:
The given expression is [tex]\(a^8 - 28a^4 + 36\)[/tex]. It is a polynomial in terms of [tex]\(a^4\)[/tex].
2. Substitution:
Let's make a substitution to simplify the expression. We will let [tex]\(u = a^4\)[/tex]. Therefore, the expression [tex]\(a^8\)[/tex] can be rewritten as [tex]\(u^2\)[/tex], since [tex]\((a^4)^2 = a^8\)[/tex].
The expression becomes:
[tex]\[ u^2 - 28u + 36 \][/tex]
3. Factoring the Quadratic Polynomial:
Now, we need to factor the quadratic polynomial [tex]\(u^2 - 28u + 36\)[/tex]. We look for two numbers that multiply to [tex]\(36\)[/tex] and add up to [tex]\(-28\)[/tex].
However, the polynomial does not factor neatly into integers, so we will solve it using other means such as completing the square or using the quadratic formula.
4. Quadratic Formula:
The quadratic formula for [tex]\(ax^2 + bx + c = 0\)[/tex] is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Applying this to [tex]\(u^2 - 28u + 36\)[/tex]:
[tex]\[ u = \frac{28 \pm \sqrt{28^2 - 4(1)(36)}}{2(1)} \][/tex]
[tex]\[ u = \frac{28 \pm \sqrt{784 - 144}}{2} \][/tex]
[tex]\[ u = \frac{28 \pm \sqrt{640}}{2} \][/tex]
[tex]\[ u = \frac{28 \pm 8\sqrt{10}}{2} \][/tex]
[tex]\[ u = 14 \pm 4\sqrt{10} \][/tex]
5. Back-Substitution:
Recall that [tex]\(u = a^4\)[/tex]. So we have two solutions for [tex]\(a^4\)[/tex]:
[tex]\[ a^4 = 14 + 4\sqrt{10} \][/tex]
[tex]\[ a^4 = 14 - 4\sqrt{10} \][/tex]
6. Final Expression:
If we were to put this back into a factorized polynomial, we can express it as:
[tex]\[ (a^4 - (14 + 4\sqrt{10}))(a^4 - (14 - 4\sqrt{10})) \][/tex]
But for simplicity and cleanliness, we recognize that the given form factors elegantly into:
[tex]\[ (a^4 - 4a^2 - 6)(a^4 + 4a^2 - 6) \][/tex]
Hence, the factored form of the expression [tex]\(a^8 - 28a^4 + 36\)[/tex] is:
[tex]\[ (a^4 - 4a^2 - 6)(a^4 + 4a^2 - 6) \][/tex]
