To solve the polynomial [tex]\(2s^4 + 13s^2 + 15 = 0\)[/tex] using u-substitution, follow these steps:
1. Define the Substitution:
Let [tex]\( u = s^2 \)[/tex]. This simplifies our polynomial from a quartic equation to a quadratic equation:
[tex]\[
2u^2 + 13u + 15 = 0
\][/tex]
2. Factor the Quadratic Polynomial:
To factor [tex]\(2u^2 + 13u + 15\)[/tex], we look for two numbers that multiply to [tex]\(2 \times 15 = 30\)[/tex] and add up to [tex]\(13\)[/tex].
These two numbers are [tex]\(10\)[/tex] and [tex]\(3\)[/tex], so we can write:
[tex]\[
2u^2 + 13u + 15 = (u + 5)(2u + 3)
\][/tex]
3. Solve for [tex]\( u \)[/tex]:
Set each factor equal to zero and solve for u:
[tex]\[
u + 5 = 0 \quad \Rightarrow \quad u = -5
\][/tex]
[tex]\[
2u + 3 = 0 \quad \Rightarrow \quad u = -\frac{3}{2}
\][/tex]
4. Substitute Back to Solve for [tex]\( s \)[/tex]:
Recall that [tex]\( u = s^2 \)[/tex]. Now solve for [tex]\( s \)[/tex] by substituting back:
[tex]\[
s^2 = -5 \quad \Rightarrow \quad s = \pm i\sqrt{5}
\][/tex]
[tex]\[
s^2 = -\frac{3}{2} \quad \Rightarrow \quad s = \pm i\sqrt{\frac{3}{2}} = \pm \frac{i\sqrt{6}}{2}
\][/tex]
5. List All Solutions:
The solutions to the original polynomial equation [tex]\( 2s^4 + 13s^2 + 15 = 0 \)[/tex] are:
[tex]\[
s = -\sqrt{5}i, \, \sqrt{5}i, \, -\frac{\sqrt{6}i}{2}, \, \frac{\sqrt{6}i}{2}
\][/tex]
Hence, the polynomial [tex]\( 2s^4 + 13s^2 + 15 = 0 \)[/tex] has the solutions:
[tex]\[
-\sqrt{5}i, \, \sqrt{5}i, \, -\frac{\sqrt{6}i}{2}, \, \frac{\sqrt{6}i}{2}
\][/tex]
