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

2 years ago

What is the volume of this solid?

olganol [36]

Answer:452.4

Step-by-step explanation:

6 0

2 years ago

How to solve the problem

WINSTONCH [101]

As tgose lines are parallel then,

a = 140° ( by alternate interior angles)

then,
tha angle just below "a" on the other paralle line would also be 140° by corresponding angles,

using exterior angle property of triangle

66° + b = 140°

b = 140° -66
= 74°

7 0

2 years ago