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Dovator [93]
3 years ago
8

How do you solve x^4 - 3x^3 - 3x^2 - 75x - 700

Mathematics
1 answer:
madam [21]3 years ago
4 0

Answer:

  = (x +4)(x -7)(x^2 +25)

  roots are -4, 7, ±5i

Step-by-step explanation:

You have not said what "solve" means in this context. An expression by itself doesn't have a solution. We have assumed you want to find the factoring and/or roots of it.

I like to use a graphing calculator to find the real roots. For this expression, there are two of them, one positive and one negative. (You know there will be one positive real root, and at least one negative real root, from Descartes's rule of signs.)

Then those roots can be factored out and the solution to the remaining quadratic determined. That factoring can occur by polynomial long division, synthetic division, or other means.

I like to see what happens when I plot the graph of the function divided by the known factors. (We expect a parabola that doesn't cross the x-axis.) The vertex of that parabola can be used to find the remaining roots.

The x-intercepts of the given expression are -4 and +7, so two of the factors are (x+4) and (x-7). Dividing these from the given expression (by synthetic division or other means) gives (x^2 +25). This only has imaginary roots (±5i).

____

If you're constrained to doing this "by hand" with only a scientific calculator, Descartes's rule of signs tells you there is one positive real root. (Only one sign change in the sequence of coefficient signs: +----.)

The rational root theorem tells you it will be a divisor of 700. Various estimates of the maximum magnitude of it will tell you it is probably less than 14 (easily checked). So, the numbers you can test as roots would be 1, 2, 4, 5, 7, 10, 14. You will find that 7 is a root, and then you can reduce the problem to the cubic x^3 +4x^2 +25x +100.

When odd-degree term signs are changed, there will be 3 sign changes (-+-+), hence at least one negative real root. The rational root theorem tells you it is a divisor of 100, so possible choices are -1, -2, -4, -5. By trial and error or other means, you can find the root to be -4. Then the problem reduces to the quadratic x^2 +25.

Roots of that are ±√(-25) = ±5i.

This process generally entails a fair amount of trial-and-error work, which is why I prefer one that makes some use of technology.

_____

We have presumed you have some familiarity with ...

  • Descartes's rule of signs
  • Rational Root Theorem
  • synthetic division

This will usually be the case when you're presented with problems like this. If you need additional information on any of these, it is readily available on the internet (and probably also in your reference material).

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Two diameters of circle P are AB and CD. Find the measure of arc ACD and arc AC when arc AD has a measure of 20 degrees
steposvetlana [31]

Answer:

Hence measure of arc ACD and arc AC are 340 degrees and 160 degrees respectively.

Step-by-step explanation:

Given:

A Circle with p as center and AB and CD as 2 diameters for a circle.

To Find:

Measure of ACD, and arc AC when arc AD is 20 degrees.

Solution:

<em>As we know that , Diameter is biggest chord which passes through center of circle,</em>

<em>Hence it divides the circle into two equal parts and making center as midpoint of it</em> .

(Refer the attachment)

angle APD=20 degree or arc AD=20 degree.

So AB as diameter ,so  arc DB=160 degree  or BPD=160 degree as A-B-P are

and angle APD and BPC are vertical opposite angles so they are congruent BPC=20 degree

Now,

points A,C,B,D,A makes circle which measures 360 degrees.

to find ACD

measure of ACD=360-(meaure of arc AD or angle APD)

=360-20

=340 degree.

Now arc AC,

as AB is diameter with p as midpoint ,A-P-B

arc APB=180 and arc BC=20 or angle BPC=20 degrees.

For measure of arc AC= measure of arc APB- arc BC

arc AC=180-20 degrees.

arc AC=160 degrees.

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3 years ago
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Alexxx [7]
The total amount of angles in quadrilateral is 360.so:

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3 years ago
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Answer:

I'm not sure, but I'll try

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Mars2501 [29]
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Let f(x)=\ln(x+1). Then f'(x)=\dfrac1{x+1}, which is positive for all x>-1, so \ln(x+1) is monotonically increasing for x>-1. This would mean \dfrac1{\ln(x+1)} must be a monotonically decreasing sequence over the same interval, and so must a_n.

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