Factor X^2-5x+6.25

[Shanel Arrendell]

Theonly really general way of which I am aware is to guess at the form of the factorization. Since it is monic (the highest term has coefficient 1), you know that the factors should also be so. Thus, there are really only 2 possible factorizations you need to think of, at least at start, which may then be further reducible through easier methods. If we denote the polynomial by $P(x)$, we produce the following candidate factorization equations:

The "obvious" next case of this would simply result in now getting a third-degree polynomial on the left and first on the right, but that's just case 1 thanks to the commutative property, so this is exhaustive. The second case is what you have here. The first case is most easily tested and solved by a simple application of the rational root theorem which will, if it's possible, give the value for $a$ - followed by a polynomial long division to get the rest.

For all real $k$ we obtain:$$x^4+10x^3+39x^2+70x+50=$$$$=(x^2+5x+k)^2-25x^2-k^2-10kx-2kx^2+39x^2+70x+50=$$$$=(x^2+5x+k)^2-((2k-14)x^2+(10k-70)x+k^2-50).$$Now, we'll choose $k$ such that we'll get a difference of squares.

For which we need $$25(k-7)^2-(2k-14)(k^2-50)=0$$ or$$(k-7)(2k^2-25k+75)=0$$ or$$(k-7)(k-5)(2k-15)=0.$$We see that only $k=7.5$ is valid and we obtain:$$$x^4+10x^3+39x^2+70x+50=(x^2+5x+7.5)^2-(x^2+5x+6.25)=$$$$=(x^2+5x+7.5)^2-(x+2.5)^2=(x^2+4x+5)(x^2+6x+10).$$

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The graph of a quadratic function is a curve called a parabola. Parabolas may open upward or downwardand vary in "width" or "steepness", but they all have the same basic "U" shape. Thepicture below shows three graphs, and they are all parabolas.

You know that two points determine a line. This means that if you are given any two points in the plane, thenthere is one and only one line that contains both points. A similar statement can be made about points and quadraticfunctions.

Given three points in the plane that have different first coordinates and do not lie on a line, there is exactlyone quadratic function f whose graph contains all three points. The applet below illustrates this fact. The graphcontains three points and a parabola that goes through all three. The corresponding function is shown in the textbox below the graph. If you drag any of the points, then the function and parabola are updated.

Sketch the graph of y = x2/2. Starting with the graph of y = x2, we shrink by a factorof one half. This means that for each point on the graph of y = x2, we draw a new point that is onehalf of the way from the x-axis to that point.

The functions in parts (a) and (b) of Exercise 1 are examples of quadratic functions in standard form.When a quadratic function is in standard form, then it is easy to sketch its graph by reflecting, shifting, andstretching/shrinking the parabola y = x2.

Any quadratic function can be rewritten in standard form by completing the square. (See the section onsolving equations algebraically to review completing the square.)The steps that we use in this section for completing the square will look a little different, because our chiefgoal here is not solving an equation.

We need to add 9 because it is the square of one half the coefficient of x, (-6/2)2 = 9. When wewere solving an equation we simply added 9 to both sides of the equation. In this setting we add and subtract 9so that we do not change the function.

In some cases completing the square is not the easiest way to find the vertex of a parabola. If the graph ofa quadratic function has two x-intercepts, then the line of symmetry is the vertical line through the midpointof the x-intercepts.

There is not much we can do with the quantity A while it is expressed as a product of two variables. However,the fact that we have only 1200 meters of fence available leads to an equation that x and y must satisfy.

We need to find the value of x that makes A as large as possible. A is a quadratic function of x, and the graphopens downward, so the highest point on the graph of A is the vertex. Since A is factored, the easiest way to findthe vertex is to find the x-intercepts and average.

After you meet certain vesting criteria, your SERS membership entitles you to receive a pension payment every month for as long as you live. The amount of the payment is determined by a formula set forth in the Retirement Code that takes into account your class of service multiplier, number of years of credited service, and final average salary.

Your SERS pension does NOT determine if or how much you may be paid for accumulated leave when you retire, offer retiree health insurance coverage, or include automatic cost of living adjustments after you retire.

The monthly payment you will receive from your SERS benefit is not influenced by how much your employer contributes or the performance of the SERS investments. It will, however, be reduced if you choose to retire before the SERS normal retirement age specified by your class of service.

Your class of service is determined largely by when you were hired and the work that you do. In most cases, your class of service will be determined by when you were first hired as a state employee. If you leave state service and later return, it is very important that you let your HR office know.

Your class of service multiplier x 2% equals your benefit accrual rate. This is one of the most common sources of confusion about SERS' pension benefits. In short hand, people will often say, "I have a 2.5% multiplier." In fact, the law provides a class of service multiplier of 1.25 that, when plugged into the formula and multiplied by 2%, yields a benefit accrual rate of 2.5%.

You are credited with one year of service when you work 1,650 hours or more in a calendar year. If you work less than 1,650 hours, you are credited with a fraction of a year equal to the number of hours you worked divided by 1,650.

In addition to earning service credit in your current job, you may be able to increase your years of credited service - and, thus, increase your pension - by purchasing service associated with time you worked in certain other jobs.

Your final average salary is the highest amount you earned during any "three non-overlapping periods of four consecutive calendar quarters." For most employees, it is the average of your last three years' salary.

Once your maximum annual retirement allowance is calculated, SERS determines how much your actual monthly payment will be based on a variety of factors, including the monthly payment option you select when you retire.

The ImmunoDiagnostics Lipocalin-2 (NGAL) ELISA is a reliable and reproducible tool for the quantitative detection of NGAL antigen. The assay is intended for use on human serum, plasma, urine, or cell culture supernatants.

NGAL Antibody Coated Microstrips: One microplate, 12 strips with 8 wells each, 96 dry wells in total. The wells are coated with a monoclonal anti- human NGAL antibody. The microplate is sealed in a foil bag. Any unused strips should be returned to the bag and resealed for future use.

Detection Antibody Solution: Concentrate (100x conc). One vial containing 0.12 mL of an HRP conjugated mouse monoclonal anti- human NGAL. Detection antibody should be diluted with only the 1X Assay Buffer needed.

Assay Buffer: Concentrate (5x conc). One vial containing 20 mL of buffer for dilution of the detection antibody. If precipitates are observed in the concentrated buffer, warm at 37oC until the precipitates disappear. Dilute 20 mL of 5x Assay Buffer with 80 mL diH2O to make a 1x concentration prior to use.

Human NGAL Standard: 25 ng of recombinant human NGAL in a buffered protein base, lyophilized. Reconstitute with 1 mL 1x Assay Buffer to prepare 25 ng/mL standard stock. Prepare a serial dilution series with 1x Assay Buffer. The reconstituted standard stock can be aliquoted and stored at -20oC for one month. Avoid repeated freeze/thaw cycles.

Wash Buffer: Concentrate (10x conc). One vial containing 50 mL of 10x wash buffer. If precipitates are observed in the 10x solution, warm at 37oC until the precipitates disappear. Prepare 1x Wash Buffer my mixing 50 mL of 10x Wash Buffer with 450 mL diH2O.

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