All categories

3 years ago

How do you factor 3u^3-4u^2-9u+12 ? Can someone please help?

Mathematics

1 answer:

Juliette [100K]3 years ago

5 0

3u³ - 4u² - 9u + 12
u²(3u) - u²(4) - 3(3u) + 3(4)
u²(3u - 4) - 3(3u - 4)
(u² - 3)(3u - 4)

You might be interested in

In the diagram shown to the right, m∠ACB=66°. Find the following.

forsale [732]

Name the symbol used in 2nd generation language

8 0

2 years ago

**IF ANYONE CAN DO ALL OF THESE QUESTIONS CORRECTLY AND FAST, YOU WILK BE MARKED BRAINLIEST!!**What is the measure of arc BC?

bixtya [17]

I think the answer is BC= (bc + de)/2

7 0

3 years ago

Very confused with this

OlgaM077 [116]

Answer:

∠ D = 28°

Step-by-step explanation:

The external angle of a triangle is equal to the sum of the 2 opposite interior angles.

∠ EBC is an exterior angle of Δ ABE, thus

∠ EBC = 90° + 34° = 124°

BD bisects ∠ EBC, so

∠ EBD = 124° ÷ 2 = 62° , then

∠ D = 180° - (90 + 62)° = 180° - 152° = 28° ( angle sum of triangle )

4 0

3 years ago

C(1, 6) ; y=4x+1<br> Put answer in slope intercept form

alina1380 [7]

Answer:

slope intercept form is y=mx+b

your m is 4

all you do is plug the numbers in

y=4x+6

4 0

3 years ago

(ASAP PICTURE ADDED) What is the simplified form of the following expression?

bonufazy [111]

Answer:

option c is correct.

Step-by-step explanation:

$7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{16x}\right)-3\left(\sqrt[3]{8x}\right)$

WE need to simplify this equation.

Solve the parenthesis of each term.

$=7\left\sqrt[3]{2x}\right-3\left\sqrt[3]{16x}\right-3\left\sqrt[3]{8x}\right$

Now, We will find factors of the terms inside the square root

factors of 2: 2

factors of 16 : 2x2x2x2

factors of 8: 2x2x2

Putting these values in our equation: $=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2X2 x}\right)-3\left(\sqrt[3]{2X2X2 x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3*2\left(\sqrt[3] {2 x}\right)-3*2\left(\sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)$

Adding like terms we get:

$=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right\\=(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)\\$

$(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)\\can\,\,be \,\, written\,\, as\,\,\\(\sqrt[3] {2x})-6\left(\sqrt[3]{x}\right)$

So, option c is correct

5 0

3 years ago

Read 2 more answers