<span><span>Solve <span>x5 + 3x4 – 23x3 – 51x2 + 94x + 120 </span></span>><span><span> 0</span>. </span></span><span>First, I factor to find the zeroes:<span><span>x5 + 3x4 – 23x3 – 51x2 + 94x + 120</span><span>= (x + 5)(x + 3)(x + 1)(x – 2)(x – 4) = 0</span></span><span>...so </span><span>x = –5, –3, –1, 2,</span><span> and </span>4<span> are the zeroes of this polynomial. (Review how to </span>solve polynomials, if you're not sure how to get this solution.)<span>To solve by the Test-Point Method, I would pick a sample point in each interval, the intervals being </span>(negative infinity, –5)<span>, </span>(–5, –3)<span>, </span>(–3, –1)<span>, </span>(–1, 2)<span>, </span>(2, 4)<span>, and </span>(4, positive infinity). As you can see, if your polynomial or rational function has many factors, the Test-Point Method can become quite time-consuming.<span>To solve by the Factor Method, I would solve each factor for its positivity: </span><span>x + 5 > 0</span><span> for </span><span>x > –5</span>;<span>x + 3 > 0</span><span> for </span><span>x > –3</span><span>; </span><span>x + 1 > 0</span><span> for </span><span>x > –1</span><span>; </span><span>x – 2 > 0</span><span> for </span><span>x > 2</span><span>; and </span><span>x – 4 > 0</span><span> for </span><span>x > 4</span>. Then I draw the grid:...and fill it in:...and solve:<span>Then the solution (remembering to include the endpoints, because this is an "or equal to" inequality) is the set of </span>x-values in the intervals<span> [–5, –3]<span>, </span>[–1, 2]<span>, and </span>[4, positive infinity]</span>. </span>
As you can see, if your polynomial or rational function has many factors, the Factor Method can be much faster.
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Answer:
It is symmetric about x- axis
The graph is not symmetric along the y - axis
It is not symmetric about the origin
Step-by-step explanation:
We can easily know the solution to your problem by graphing the equation
r(θ) = 2*cos (5*θ)
Please, see attached image
The graph is symmetric about the x axis if
r(θ) = r(-θ)
r(θ) = 2*cos (5*θ) = 2*cos (-5*θ) = r(-θ)
Since the cosine is an even function.
It is symmetric about x- axis
The graph is not symmetric along the y - axis, because
r(θ) ≠ r(π-θ)
It is symmetric about the origin if
r(θ) = r(π+θ)
But, since,
r(θ) = 2*cos (5*θ) ≠ 2*cos (5*(θ+π)) = 2*cos (5*θ+5*π)
It is not symmetric about the origin
