I'd like to solve question number 9 from practice quiz week 2.

And I also want to let you know that I have solved a similar problem.

On Vaughan Videos found within text section 3.5, number 11 is a similar problem.

Let's solve. We have polynomial function of P(x)=x^4+6x^2+9.

Rewriting as 0=x^4+6x^2+9 to solve.

x^4+6x^2+9=0.

Let me remind you of a former topic, This function is a "quadratic in form" function; in other words it's a perfect square trinomial many of you may recall from your studies in M99 or Algebra 2.

Let's review one solving technique for a perfect trinomial quadratic in from function.

We have a fourth degree polynomial in which the first term, x^4 is a perfect square and the constant, 9, is also a perfect square, 3. Let's solve this using a "U" substitution.

Let x^2=U.

Directly substitute in U for the x^2 term as follows, u^2+6u+9=0 u^2+6u+9=0 u^2+6u+9=0 u^2+6u+9=0 Now, solve for the perfect square by factoring as follows,

(u+3)(u+3)=0

is factored.

You could FOIL back out to justify.

Justifying.

Justifying.

Justifying.

Justifying.

Justifying. Is 9.

Remember that a perfect square trinomial follows the pattern of a^2+2ab+b^2.

This expands to our original eqn.

Ok, continuing on.

Solve for (u+3)(u+3)=0 Solving, (u+3)(u+3)=0

u+3=0 Thus, u=-3, however, Recall that we substituted u for x^2.

now, replace the u with x^2.

Thus, x^2=-3 x^2=-3 Solving, take a sq rt of both sides, leaving x= +-sqrt(-3) x= +-sqrt(-3) x=+-i*sqrt(3) x=+-i*sqrt(3) Reviewing taking the sq rt of (-1) √ -1 =i Final answer, x=+-i sqrt(3) with a multiplicity of 2 as this is a fourth degree polynomial.

And, I hope that helps.

This is your final answer.

And, this is called a 4th degree polynomial that is quadratic in form.