Question

1. which of the following are zeros of \\( p(x) = x^5 - 11x^3 + 18x \\)?

Explanation:

Step1: Factor the polynomial

To find the zeros of \( P(x) = x^5 - 11x^3 + 18x \), first factor out the common factor \( x \):
\( P(x)=x(x^4 - 11x^2 + 18) \)
Now, let \( y = x^2 \), then the quadratic in \( y \) is \( y^2 - 11y + 18 \). Factor this quadratic:
We need two numbers that multiply to \( 18 \) and add to \( - 11 \). The numbers are \( -2 \) and \( -9 \). So,
\( y^2 - 11y + 18=(y - 2)(y - 9) \)
Substitute back \( y = x^2 \):
\( x^4 - 11x^2 + 18=(x^2 - 2)(x^2 - 9) \)
And \( x^2 - 9 \) is a difference of squares, so \( x^2 - 9=(x - 3)(x + 3) \), and \( x^2 - 2=(x-\sqrt{2})(x + \sqrt{2}) \)

Step2: Combine the factors

Putting it all together, the factored form of \( P(x) \) is:
\( P(x)=x(x - \sqrt{2})(x + \sqrt{2})(x - 3)(x + 3) \)

Step3: Find the zeros

To find the zeros, set \( P(x) = 0 \):
\( x(x - \sqrt{2})(x + \sqrt{2})(x - 3)(x + 3)=0 \)
Using the zero - product property (if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \)), we get:
\( x = 0 \) or \( x-\sqrt{2}=0\Rightarrow x = \sqrt{2} \) or \( x+\sqrt{2}=0\Rightarrow x=-\sqrt{2} \) or \( x - 3=0\Rightarrow x = 3 \) or \( x + 3=0\Rightarrow x=-3 \)

So the zeros of \( P(x) \) are \( x = 0,x=\sqrt{2},x = -\sqrt{2},x = 3,x=-3 \)

Answer:

The zeros of \( P(x)=x^{5}-11x^{3}+18x \) are \( \boldsymbol{-3,-\sqrt{2},0,\sqrt{2},3} \)