All categories

3 years ago

We want to find the zeros of this polynomial: p(x)= 3x^3 – 3x^2– 18x

Mathematics

2 answers:

bulgar [2K]3 years ago

3 0

Answer:

x=0 x=3 x=-2

Step-by-step explanation:

p(x)= 3x^3 – 3x^2– 18x

Factor out the greatest common factor, 3x

p(x)= 3x (x^2 – x– 6)

Factor inside the parentheses

What 2 numbers multiplies to -6 and adds to -1

-3*2 = -6

-3+2 = -1

p(x)= 3x (x-3)(x+2)

Setting the function equal to zero to find the zeros

0 = 3x (x-3)(x+2)

Using the zero product property

3x = 0 x-3 =0 x+2 =0

x=0 x=3 x=-2

madam [21]3 years ago

3 0

Answer:

x = -2, 0, 3.

Step-by-step explanation:

3x^3 – 3x^2– 18x = 0

First take out the GCF which is 3x:

3x(x^2 - x - 6) = 0

So one zero is 0 ( because 3x = 0).

x^2 - x - 6 = 0

(x - 3)(x + 2) = 0

x = -2, 3.

You might be interested in

The blueprint for a new swimming pool uses the scale 3/4 inch = 2 feet. If the actual dimensions of the pool will be 8 ft wide b

weeeeeb [17]

Answer:

The dimensions of the pool on the blueprint would be 3 in wide by 4.5 in long.

Step-by-step explanation:

First I turned 3/4 into a decimal, which would be 0.75. The I divided the dimensions of the pool by two. This way I could multiply the .75 by how many sets of two feet there were in the dimensions. Finally I multiplied 4 by .75 and I multiplied 6 by .75. That is how I found my answer.

5 0

3 years ago

Given:• PQRS is a rectangle.• mZ1 = 50°Р21SRWhat is mZ2?130°85°70°65°

Over [174]

To answer this question, we need to recall that: "the diagonals of a rectangle bisect each other"

Thus, if we assign the point of intersection of the two diagonals in the rectangle as point O, we can say that the triangle OQR is an "isosceles triangle". Note that this is because the lengths OR and OQ are equal since we know that: "the diagonals of a rectangle bisect each other". See the below diagram for clarity.

Now, we have to recall that:

- the base angles of any isosceles triangle are equal. This is a fact, and this means that the angles

- also the sum of all the angles in any triangle is 180 degrees

Now, considering the isosceles triangle OQR, we have that:

$\angle OQR+\angle ORQ+\angle ROQ=180^o$

Now, since the figure already shows that angle $m\angle2+\angle ORQ+50^o=180^o$ Now, since we have established that the base angles $m\angle2+m\angle2+50^o=180^o$ we can now solve the above equation for m<2 as follows:

$\begin{gathered} m\angle2+m\angle2+50^o=180^o \\ \Rightarrow2m\angle2+50^o=180^o \\ \Rightarrow2m\angle2=180^o-50^o \\ \Rightarrow2m\angle2=130^o \\ \Rightarrow m\angle2=\frac{130^o}{2}=65^o \end{gathered}$

Therefore, the correct answer is: option D