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3 years ago

What is the completely factored form of this polynomial? 18x^3 – 120x^2 – 42x

Mathematics

2 answers:

Nookie1986 [14]3 years ago

8 0

The completely factored form is 6x(3x+1)(x-7)

lyudmila [28]3 years ago

4 0

Answer:

6x(3x + 1)(x - 7) .

Step-by-step explanation:

First look for the greatest common factor of the 3 terms.

It is 6x, so

18x^3 – 120x^2 – 42x

= 6x(3x^2 - 20x - 7)

Now we look to factor the quadratic expression in the parentheses:

Use the 'ac' method:

The first and last coefficients are multiplied

3 * -7 = -21.

We look for 2 numbers whose product is -21 and whose sum is -20:

They are -21 and + 1 , so we write:

6x(3x^2 - 21x + 1x - 7)

= 6x [ (3x(x - 7) + 1(x - 7)] Note that x - 7 is common, so:

= 6x(3x + 1)(x - 7) Answer

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sukhopar [10]

Answer:

4^6

Step-by-step explanation:

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3 years ago

A. 8<br> b. 10<br> c. 3<br> d. 4

NeTakaya

Answer:

Step-by-step explanation:

The number on the right squared in equal to the number on the left

5^2 =25

6^2 = 36

What squared = 16

4^2 = 16

The the missing number is 4

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3 years ago

7 less than a number is 20

sesenic [268]

Answer:

7 - x = 20 hope this helps!!

Step-by-step explanation:

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3 years ago

Read 2 more answers

2m - 6 = 8m then 3m =

Vinil7 [7]

Answer:

3m = -3

Step-by-step explanation:

You are given

2m - 6 = 8m,

hence,

2m - 8m = 6,

-6m = 6,

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m = -1.

Therefore, 3m = -3.

5 0

3 years ago

30 POINTS AVAILABLE

kherson [118]

Answer:

$\large\boxed{(x-2)^2+(y-1)^2=34}$

Step-by-step explanation:

The equation of a circle in standard form:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

We have the endpoints of the diameter: (-1, 6) and (5, -4).

Midpoint of diameter is a center of a circle.

The formula of a midpoint:

$\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)$

Substitute:

$h=\dfrac{-1+5}{2}=\dfrac{4}{2}=2\\\\k=\dfrac{6+(-4)}{2}=\dfrac{2}{2}=1$

The center is in (2, 1).

The radius length is equal to the distance between the center of the circle and the endpoint of the diameter.

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the coordinates of the points (2, 1) and (5, -4):

$r=\sqrt{(5-2)^2+(-4-1)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}$

Finally we have:

$(x-2)^2+(y-1)^2=(\sqrt{34})^2$

3 0

3 years ago