Answer :

To factor the polynomial
[tex]$$
x^4 - 12x^2 - 45,
$$[/tex]
we can use a substitution to simplify the process.

1. Let
[tex]$$
u = x^2.
$$[/tex]
Then [tex]$x^4$[/tex] becomes [tex]$u^2$[/tex], and the polynomial transforms into a quadratic in [tex]$u$[/tex]:
[tex]$$
u^2 - 12u - 45.
$$[/tex]

2. To factor the quadratic [tex]$u^2 - 12u - 45$[/tex], compute the discriminant:
[tex]$$
D = (-12)^2 + 4 \cdot 45 = 144 + 180 = 324.
$$[/tex]
The square root of the discriminant is
[tex]$$
\sqrt{324} = 18.
$$[/tex]

3. Next, find the roots of the quadratic using the quadratic formula:
[tex]$$
u = \frac{12 \pm 18}{2}.
$$[/tex]
This yields the two solutions:
[tex]$$
u_1 = \frac{12 + 18}{2} = 15 \quad \text{and} \quad u_2 = \frac{12 - 18}{2} = -3.
$$[/tex]

4. The quadratic factors accordingly as:
[tex]$$
u^2 - 12u - 45 = (u - 15)(u + 3).
$$[/tex]

5. Finally, substitute [tex]$u = x^2$[/tex] back into the factors:
[tex]$$
(u - 15)(u + 3) = (x^2 - 15)(x^2 + 3).
$$[/tex]

Thus, the complete factorization of the polynomial is:
[tex]$$
(x^2 - 15)(x^2 + 3).
$$[/tex]

Thanks for taking the time to read Factor using substitution If the polynomial is prime enter PRIME tex x 4 12x 2 45 tex. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada