A polynomial of degree 3 has exactly 3 solutions if we use complex numbers. If the polynomial has real coefficients and z is a complex solution, i.e., $p(z)=0$ , the conjugate is also a solution: $p(\overline{z})=0$

So, in the first case, if we know that 1-i is a solution, the conjugate 1+i must be a solution as well.

So, we know all the roots: -5, 1+i, 1-i.

When you know the solutions $x_1,\ x_2,\ldots,\ x_n$ of a polynomial, you can write the polynomial (up to multiples) as

$p=(x-x_1)(x-x_2)\ldots(x-x_n)$

So, in your case, we have

$p(x) = (x+5)(x-1-i)(x-1+i) = x^3+3x^2-8x+10$

You can check that indeed $p(0)=10$ , because the constant term is 10, so, we're ok with this

The second exercise is exactly the same: the solutions -6, i, -i identify the polynomial

$p(x)=(x+6)(x-i)(x+i)=x^3+6x^2+x+6$

In this case, P(-3)=30, so if we want P(-3)=60 we have to multiply everything by 2:

$p(x)=2x^3+12x^2+2x+12$

3 0

3 years ago

Consider the following data set: <br> 11, 8, 0, 8, 13, 3, 14 <br> Find the median.

r-ruslan [8.4K]

Answer:

The median is 8.

How you get that is that you have to order these from smallest to biggest.

0,3,8,8,11,13,14

Next, you cut one from each side until you get one single number left. If you get two, then you add them together and divide it by 2.

3 0

3 years ago

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Solve the inequality <br> V/3+12>7

Ivanshal [37]

Answer:

V > -15

Step-by-step explanation:

$\frac{V}{3} +12>7\\\\\frac{V}{3}>-5\\\\V>-15$

7 0

3 years ago

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