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Featured 6 months ago

According to Factor theorem, for a polynomial

Hence, if we have a polynomial say

Observe that coefficient of highest power of

In practice, Factor Theorem is used for factorising polynomials completely. So our steps in factorising a polynomial

1 Check the constant term in

2. Find its all possible factors.

3. Take one of the factors, say

4. Try more factors whose number should be equal to the degree of polynomial.

And you have got all the factors.

Coming to your question, we have

or

or

This is the same as in

We should have zeros among

Note that it is easier to check whether

Featured 5 months ago

See below.

First graph the line

With graph plotted, there will be four possible regions: **A**, **B** , **C** , **D**

We have to test coordinates in each region to identify which are included regions and which are excluded regions:

Region **A**

**FALSE**

**Region A is an excluded region.**

Region **B**

**FALSE**

**Region B is an excluded region.
**

Region

**TRUE**

**Region C is an included region.
**

Region **D**

**TRUE**

**Region D is an included region.**

Shade included regions **C** and **D**

Featured 5 months ago

Determination of the zeros and a sign chart of

**Finding the zeros of #f(x)=x^3(x+2)^2#**

Remember that a function has the value zero at (and only at) points where some factor of the function has the value zero.

The factors of

are

and the only unique factors are

Therefore

For non-zero values of

We can pick arbitrary values within each interval to determine if

**Unfortunately this does **not** tell us any detailed information about the behavior of

Just for reference, here is what the graph should look like, but you would need to use something beyond the zeros and a sign chart to sketch this.

Featured 2 months ago

A few more...

**Elementary symmetric polynomials**

The coefficients of a monic polynomial are (modulo alternating signs), the elementary symmetric polynomials in the zeros.

For example:

#(x-alpha)(x-beta) = x^2-(alpha+beta)x+alphabeta#

#(x-alpha)(x-beta)(x-gamma) = x^3-(alpha+beta+gamma)x^2+(alphabeta+betagamma+gammaalpha)x-alphabetagamma#

etc.

In particular, given a polynomial:

#a_n x_n + a_(n-1) x_(n-1) + ... + a_1 x + a_0#

the sum of its zeros is

Since any symmetric polynomial can be constructed from the elementary symmetric polynomials, we can find a polynomial that has zeros that are (say) the squares of the zeros of another polynomial - without finding what the zeros actually are.

For a substantial and important application of this, see https://socratic.org/s/aPGxwybx

**Coefficient sum shortcuts**

If the sum of the coefficients of a polynomial is

If inverting the signs on terms of odd degree results in coefficients that sum to

**Reversing the order and reciprocals**

Given a polynomial:

#a_n x_n + a_(n-1) x^(n-1) + ... + a_1 x + a_0#

with

Then the polynomial:

#a_0 x_n + a_1 x_(n-1) + ... + a_(n-1) x + a_n#

has zeros which are reciprocals of the original polynomial.

To see why that is so, note that:

#1/x_n (a_n x_n + a_(n-1) x^(n-1) + ... + a_1 x + a_0)#

#=a_n + a_(n-1) 1/x + ... + a_1 1/x^(n-1) + a_0 1/x^n#

**Symmetry and reciprocals**

From the preceding property, we can deduce that:

If the coefficients of a polynomial are symmetrical then you can infer that the reciprocal of any zero is also a zero.

For example:

#6x^4+5x^3-38x^2+5x+6#

has zeros

Featured 2 months ago

Let us locate the outer corner of the corridor at

Consider lines of negative slope passing through

The minimum distance then models the maximum feasible length of beam.

graph{((x-14)^100+(y-13)^100-10^100)((x-20)^100+(y-20)^100-20^100)(y+x/2-5)((x-4)^2+(y-3)^2-0.02)(x^2+(y-5)^2-0.02)((x-10)^2+y^2-0.02) = 0 [-5.625, 14.375, -3, 7]}

The equation of such a line can be written in point slope form as:

#y-b = m(x-a)#

The

#x = a-b/m#

and the

#y = b-am#

The square of the distance between these intercepts is:

#(a-b/m)^2+(b-am)^2 = a^2-2ab 1/m + b^2 1/m^2 + b^2-2ab m + a^2 m^2#

The minimum will occur when the derivative of this with respect to

#0 = 2ab 1/m^2 - 2b^2 1/m^3-2ab+2a^2m#

Multiplying by

#0 = abm-b^2-abm^3+a^2m^4#

#color(white)(0) = a^2m^4-abm^3+abm-b^2#

#color(white)(0) = am^3(am-b)+b(am-b)#

#color(white)(0) = (am^3+b)(am-b)#

#color(white)(0) = (a^(1/3)m+b^(1/3))(a^(2/3)m^2-a^(1/3)b^(1/3)m+b^(2/3))(am-b)#

Note here that the last linear factor gives

Also the quadratic factor has only non-real solutions.

So we require

With this value of

#(a-b/m)^2+(b-am)^2 = (a+a^(1/3)b^(2/3))^2+(b+a^(2/3)b^(1/3))^2#

#color(white)((a-b/m)^2+(b-am)^2) = a^(2/3)(a^(2/3)+b^(2/3))^2+b^(2/3)(b^(2/3)+a^(2/3))^2#

#color(white)((a-b/m)^2+(b-am)^2) = (a^(2/3)+b^(2/3))^3#

So the distance is:

#sqrt((a^(2/3)+b^(2/3))^3) = (a^(2/3)+b^(2/3))^(3/2)#

Featured 3 weeks ago

Create a **data table** to graph the **exponential function : **

Create a **data table**:

Closely examine the values for

No value of

So, the graph gets closer and closer to the **x-axis** but never touches the **x-axis.**

Hence, **horizontal asymptote**.

Using the data table, graph the function

For easy comprehension, graph the exponential function **parent function**.

Compare both the graphs to understand the behavior of the given exponential function

The graph of **reflected about the x-axis ** when the sign of **negative**.

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