The equation X^3+3x^2 + 5x+15 = 0 Has how many solutions

1 answer:

8 0

x
=
-
3
,
5
,
-
5
The result can be shown in multiple forms.
Exact Form:
x
=
−
3
,
√
5
,
−
√
5
x
=
-
3
,
5
,
-
5
Decimal Form:
x
=
−
3
,
2.23606797
…
,
−
2.23606797
…

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Help me or give me answer key plz asap

Brut [27]

Hello, this should help you out!

In this graph, the y values move up 3 units each. Therefore, the blue parabola underwent a translation of 3 units upward.

For further explanation, we see that (0,0) goes up to (0,3), meaning the graph shifted upwards 3 units. So, each y value is raised by positive 3 units!

I hope I was of assistance! #SpreadTheLove! <3

3 0

3 years ago

Which is the following is the set of real zeros of the function f(x) = (x3 + 1000)(x4 - 160,000)?

marusya05 [52]

Answer:

A. { -20, -10, 20 }

Step-by-step explanation:

Given:

The function is given as:

$f(x)=(x^3+1000)(x^4-160000)$

Let us simplify the function.

First, we use the identity $a^3+b^3=(a+b)(a^2-ab+b^2)$

$x^2+1000= x^3+10^3=(x+10)(x^2-10x+10^2)\\x^3+1000=(x+10)(x^2-10x+100)$

Next, we use the identity $a^4-b^4=(a-b)(a+b)(a^2+b^2)$

$x^4-160000=x^4-20^4=(x-20)(x+20)(x^2+20^2)=(x-20)(x+20)(x^2+400)$

Now, the function can be rewritten as:

$f(x)=(x+10)(x^2-10x+100)(x-20)(x+20)(x^2+400)$

Now, the zeros are those values of $x$ for which $f(x)=0$

Now, for $f(x)=0$ , we must have either of the factors 0.

$x+10=0\\x=-10$

$x-20=0\\x=20\\\\x+20=0\\x=-20$

The factors $x^2-10x+100$ and $x^2+400$ can have no zeros as the first one has imaginary roots and second one is always greater than 0 irrespective of the $x$ values.