Find the zeros of the following
A) p(x)= x²-6x²-16x
There seems to be a typographical error in the given polynomial. It should be either "p(x) = x² - 6x + 16x" or "p(x) = x² - 6x - 16x". Assuming the former case:
p(x) = x² - 6x + 16x
= x² + 10x
= x(x + 10)
Therefore, the zeros of p(x) are x = 0 and x = -10.
To find the zeros of the polynomial \(p(x) = x^2 - 6x^2 - 16x\), we need to solve the equation \(p(x) = 0\).
First, let's rewrite the equation:
\[x^2 - 6x^2 - 16x = 0\]
Combining like terms:
\[-5x^2 - 16x = 0\]
Now, let's factor out the common factor of \(x\):
\[x(-5x - 16) = 0\]
Since the product of two numbers is zero if and only if at least one of the numbers is zero, we have two possibilities:
1) \(x = 0\)
2) \(-5x - 16 = 0\)
Solving the second equation:
\[-5x - 16 = 0\]
Add 16 to both sides:
\[-5x = 16\]
Divide by -5:
\[x = \frac{16}{-5}\]
\[x = -\frac{16}{5}\]
Therefore, the zeros of the polynomial \(p(x) = x^2 - 6x^2 - 16x\) are \(x = 0\) and \(x = -\frac{16}{5}\).