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3 years ago

Find the value of c so that (x+2) is a factor of the polynomial p(x)=3x^4-x^2+cx-2

Mathematics

1 answer:

DedPeter [7]3 years ago

4 0

Using polynomial long division, we get

3x^3+6x^2+11x
_____________
(x+2) | 3x^4-x^2+cx-2
-(3x^4+6x^3)
____________
6x^3-x^2+cx-2
- (6x^3+12x^2)
_____________
11x^2+cx-2
-(11x^2+22x)
__________
(22+c)x-2.

If you're wondering how I did the long division, what I essentially did was get the first value (at the start, it was 3x^4) and divided it by the first value of the divisor (which in x+2 was x) to get 3x^3 in our example. I then subtracted the polynomial by the whole divisor multiplied by, for example, 3x^3 and repeated the process.

For this to be a perfect factor, (x+2)*something must be equal to (22+c)x-2. Focusing on how to cancel out the 2, we have to add 2 to it. To add 2 to it, we have to multiply (x+2) by 1. However, there's a catch, which is that we subtract whatever we multiply (x+2) by, so we have to multiply it by -1 instead. We still need to cross out (22+c)x. Multiplying (x+2) by -1, we get
(-x-2) but by subtracting the whole thing from something means that we have to add -(-x-2)=x+2 to something to get 0. x+2-x-2=0, xo (22+c)x-2 must equal -x-2, meaning that (22+c)=-1 and c=-23

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Write the equation of the parabola that has its x-intercepts at (1+<img src="https://tex.z-dn.net/?f=%5Csqrt%7B5%7D" id="TexForm

VladimirAG [237]

Answer:

$y=2x^2-4x-8$

Step-by-step explanation:

<u>Factored form of a parabola</u>

$y=a(x-p)(x-q)$

where:

p and q are the x-intercepts.
a is some constant.

Given x-intercepts:

(1+√5, 0)
(1-√5, 0)

Therefore:

$\implies y=a(x-(1+\sqrt{5}))(x-(1-\sqrt{5}))$

$\implies y=a(x-1-\sqrt{5})(x-1+\sqrt{5})$

To find a, substitute the given point (4, 8) into the equation and solve for a:

$\implies a(4-1-\sqrt{5})(4-1+\sqrt{5})=8$

$\implies a(3-\sqrt{5})(3+\sqrt{5})=8$

$\implies4a=8$

$\implies a=2$

Therefore, the equation of the parabola in factored form is:

$\implies y=2(x-1-\sqrt{5})(x-1+\sqrt{5})$

Expand so that the equation is in standard form:

$\implies y=2(x^2-x+\sqrt{5}x-x+1-\sqrt{5}-\sqrt{5}x+\sqrt{5}-5)$

$\implies y=2(x^2-x-x+\sqrt{5}x-\sqrt{5}x+\sqrt{5}-\sqrt{5}+1-5)$

$\implies y=2(x^2-2x-4)$

$\implies y=2x^2-4x-8$

6 0

1 year ago

Blood pressure values are often reported to the nearest 5 mmHg (100, 105, 110, etc.). The actual blood pressure values for nine

Paladinen [302]

Answer:

a) 117.4

b) 117.9

c) Option A) When there is rounding or grouping, the median can be highly sensitive to small change

Step-by-step explanation:

We are given the following data set in the question:

108.6, 117.4, 128.4, 120.0, 103.7, 112.0, 98.3, 121.5, 123.2

$Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}$

n = 9

a) Median of the reported blood pressure values

Sorted Values: 98.3, 103.7, 108.6, 112.0, 117.4, 120.0, 121.5, 123.2, 128.4

Median =

$\dfrac{9 + 1}{2}^{th}\text{ term} = 5^{th}\text{ term} = 117.4$

b) New median of the reported values

Data: 108.6, 117.9, 128.4, 120.0, 103.7, 112.0, 98.3, 121.5, 123.2

Sorted Values: 98.3, 103.7, 108.6, 112.0, 117.9, 120.0, 121.5, 123.2, 128.4

New Median =

$\dfrac{9 + 1}{2}^{th}\text{ term} = 5^{th}\text{ term} = 117.9$

c) Since median is a position based descriptive statistics, a small change in values can bring a change in the median value as the order of the data may change.

Option A) When there is rounding or grouping, the median can be highly sensitive to small change

7 0

3 years ago

Caleb bought groceries and paid

lutik1710 [3]

Answer:

Price of Caleb's groceries before tax= $6.4

Step-by-step explanation:

Let price of groceries before tax be = $ $x$