Can someone please help me with this, and please only comment if ur sure, Thank you <3

Write the polynomial f(x)=x^4-10x^3+25x^2-40x+84. In factored form

Verizon [17]

<h2>Steps:</h2>

So firstly, to factor this we need to first find the potential roots of this polynomial. To find it, the equation is $\pm \frac{p}{q}$ , with p = the factors of the constant and q = the factors of the leading coefficient. In this case:

$\textsf{leading coefficient = 1, constant = 84}\\\\p=1,2,3,4,6,7,12,14,21,28,42,84\\q=1\\\\\pm \frac{1,2,3,4,6,7,12,14,21,28,42,84}{1}\\\\\textsf{Potential roots =}\pm 1, \pm 2,\pm 3,\pm 4,\pm 6, \pm 7,\pm 12,\pm 14,\pm 21,\pm 28,\pm 42,\pm 84$

Next, plug in the potential roots into x of the equation until one of them ends with a result of 0:

$f(1)=(1)^4-10(1)^3+25(1)^2-40(1)+84\\f(1)=1-10+25-40+84\\f(1)=60\ \textsf{Not a root}\\\\f(2)=2^4-10(2)^3+25(2)^2-40(2)+84\\f(2)=16-10*8+25*4-80+84\\f(2)=16-80+100-80+84\\f(2)=80\ \textsf{Not a root}\\\\f(3)=3^4-10(3)^3+25(3)^2-40(3)+84\\f(3)=81-10*27+25*9-120+84\\f(3)=81-270+225-120+84\\f(3)=0\ \textsf{Is a root}$

Since we know that 3 is a root, this means that one of the factors is (x - 3). Now that we know one of the roots, we are going to use synthetic division to divide the polynomial. To set it up, place the root of the divisor, in this case 3 from x - 3, on the left side and the coefficients of the original polynomial on the right side as such:

3 | 1 - 10 + 25 - 40 + 84
_________________

Firstly, drop the 1:

3 | 1 - 10 + 25 - 40 + 84
↓
_________________
1

Next, multiply 3 and 1, then add the product with -10:

3 | 1 - 10 + 25 - 40 + 84
↓ + 3
_________________
1 - 7

Next, multiply 3 and -7, then add the product with 25:

3 | 1 - 10 + 25 - 40 + 84
↓ + 3 - 21
_________________
1 - 7 + 4

Next, multiply 3 and 4, then add the product with -40:

3 | 1 - 10 + 25 - 40 + 84
↓ + 3 - 21 + 12
_________________
1 - 7 + 4 - 28

Lastly, multiply -28 and 3, then add the product with 84:

3 | 1 - 10 + 25 - 40 + 84
↓ + 3 - 21 + 12 - 84
_________________
1 - 7 + 4 - 28 + 0

Now our synthetic division is complete. Now since the degree of the original polynomial is 4, this means our quotient has a degree of 3 and follows the format $ax^3+bx^2+cx+d$ . In this case, our quotient is $x^3-7x^2+4x-28$ .

So right now, our equation looks like this:

$f(x)=(x-3)(x^3-7x^2+4x-28)$

However, our second factor can be further simplified. For the second factor, I will be factoring by grouping. So factor x³ - 7x² and 4x - 28 separately. Make sure that they have the same quantity inside the parentheses:

$f(x)=(x-3)(x^2(x-7)+4(x-7))$

Now it can be rewritten as:

$f(x)=(x-3)(x^2+4)(x-7)$

<h2>Answer:</h2>

Since the polynomial cannot be further simplified, your answer is:

$f(x)=(x-3)(x^2+4)(x-7)$

6 0

3 years ago

Diana can use the equation y = 7x to calculate her pay, where y represents the amount of pay, and x represents the number of hou

kherson [118]

You have to take the amount of money payed, which in this case is 45.50, and divide it with 7.

8 0

3 years ago

Read 2 more answers

In 2000 the number of students enrolled at Arlington County day school was 823 students in 2010 the population was approximately

RoseWind [281]

Answer:

enrollment decreases by 14.34%.

Step-by-step explanation:

Given that number of students enrolled in 2000 = 823

Given that number of students enrolled in 2010 = 705

705 is less than 823 so that means number of students enrollment is decreasing.

Decrease = 823-705 = 118

Percent decrease is given by formula:

$Percent =\frac{part}{whole}\cdot100 =\frac{118}{823}\cdot100= 14.3377885784$

which is approx 14.34%.

Hence final answer is that enrollment decreases by 14.34%

5 0

3 years ago

Steven wants to buy a new video game. He needs at least $48. He has $26.00 saved. His brother said he would pay him $5 an hour t

Law Incorporation [45]

I think that the answer is C

5 0

3 years ago

A sine function is transformed such that it has a single x-intercept in the interval (0,pi), a period of pi and a y-intercept of

Alex787 [66]

Just to make sure we're using the same language, I'm going to use the function form of:

$y = A\sin(kx) + h$

[] I would agree that k = 2, since the period is only half as long as a normal sine function. So, we so far, have y = A sin(2x) + h. We still need to find A and h.

[] The y-intercept is 3. Remember that the y-intercept happens when x = 0. So, plugging in x = 0 into our formula, we have: 3 = A sin(2*0) + h. In other words, 3 = A sin(0) + h = 0 + h = h. So, we now know that h = 3. The formula is now y = A sin(2x) + 3.

[] Finally, there is a single x-intercept. Picture what the sine function looks like right now, it is floating in the air around y = 3. We need to stretch it vertically until it just grazes the x-axis.

If A = 1, then our sine function bounces between 2 and 4 (+/- 1 around h = 3). But that doesn't touch 0, so no good.

If A = 2, then our sine function bounces between 1 and 5 (+/- 2 around h = 3). Again, not quite touching 0 yet.

The answer should be A = 3, then our sine function bounces between 0 and 6 (+/- 3 around h = 3).

The final formula is y = 3 sin(2x) + 3.

7 0

3 years ago