MATHEMATICS HIGH SCHOOL

Find the zeros of the following polynomial.
3x3 + 9x2 - 12x
Find the zeros of the following polynomial. 3x3 + 9x2 - 1

Answers

Answer 1
Answer: x = 4
x = -1
x = 0

Equation at the end of step 1 :

((3 • (x3)) - 32x2) - 12x = 0
Step 2 :

Equation at the end of step 2 :

(3x3 - 32x2) - 12x = 0
Step 3 :

Step 4 :

Pulling out like terms :

4.1 Pull out like factors :

3x3 - 9x2 - 12x = 3x • (x2 - 3x - 4)

Trying to factor by splitting the middle term

4.2 Factoring x2 - 3x - 4

The first term is, x2 its coefficient is 1 .
The middle term is, -3x its coefficient is -3 .
The last term, "the constant", is -4

Step-1 : Multiply the coefficient of the first term by the constant 1 • -4 = -4

Step-2 : Find two factors of -4 whose sum equals the coefficient of the middle term, which is -3 .

-4 + 1 = -3 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -4 and 1
x2 - 4x + 1x - 4

Step-4 : Add up the first 2 terms, pulling out like factors :
x • (x-4)
Add up the last 2 terms, pulling out common factors :
1 • (x-4)
Step-5 : Add up the four terms of step 4 :
(x+1) • (x-4)
Which is the desired factorization
Answer 2
Answer:

If you factor 3x from the expression, you have

3x^3+9x^2-12x=3x(x^2+3x-4)

So, we have

3x(x^2+3x-4)=0 \iff 3x=0\lor x^2+3x-4=0

We easily have

3x=0\iff x=0

So, one solution is x=0.

The other solutions depend on the quadratic equation:

x^2+3x-4=0 \iff x=-4 \lor x=1

So, the solutions are x=-4,\ 0,\ 1

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MATH HELP MIDPOINT…..

Guys I’ve got 15 minutes to answer all of them please help I’ve been busy with the rest of the problems help !!!!

Answers

Answer:

Step-by-step explanation:

yeah on the first on i got x= 54 but my brain is mush to do the rest sorry my dude

Somebody help (no links please) this is stressing me rn cause this assignment is almost due.

Answers

Step-by-step explanation:

11)

d=√[(-2-(-3))^2+(3-10)^2]=√(-2+3)^2+(-7)^2

=√1+49=√50=7.07 approximately

12)

M (((-1)+(-7))/2 , (2+(-4))/2)=(-8/2 , -2/2)=(-4 ,-1)

What is the value of x in the equation 3x – 4y = 65, when y = 4?

Answers

27 because 4*4=16. 65+16=81 and 81 divided by 3=27

At a local hospital, 35 babies were born. if 23,were boys, what percentage of the newborns were boys?

Answers

The percentage of newborns who were boys is 65.71%.

Given that

At a local hospital, 35 babies were born.

There are 23 were boys.

We have to determine

What percentage of the newborns were boys?

According to the question

At a local hospital, 35 babies were born.

There are 23 were boys.

Then,

The percentage of the newborns were boys is determined by,

\rm Percentage \ of \ new \ born \ boys = (Total \ number\ of \ boys * 100)/(Total \ number \ of \ babies)

Substitute all the values in the formula.

\rm Percentage \ of \ new \ born \ boys = (Total \ number\ of \ boys * 100)/(Total \ number \ of \ babies)\n\n\rm Percentage \ of \ new \ born \ boys = (23 * 100)/(35)\n\n\rm Percentage \ of \ new \ born \ boys = (2300)/(35)\n\n\rm Percentage \ of \ new \ born \ boys = 65.71 \ percent

Hence, the percentage of newborns who were boys is 65.71%.

To know more about Percentages click the link given below.

brainly.com/question/8009466
so the fraction would be 23/35 are boys. If you do the division (23 ÷ 35) the answer is around 0.657.

That converted into a percentage (×100) is 65.7%

Please help me with this math question?? :)The measure of each exterior angle of a regular octogon is _____ the measure of each exterior agle of a regular hexagon.

A. Greater than
B. Less than
C. Equal to

Answers

\bf \textit{exterior angle of a regular polygon}\n\n\n\theta=\cfrac{360}{n}\qquad \begin{cases}\theta=\textit{the angle in degrees}\nn=\textit{number of sides}\end{cases}

now, an OCTAgon, has OCTA=8, sides
an HEXAgon, has HEXA=6, sides

check what their external angles are, and compare :)

Expresa en forma algebraica. ⦁ El cuadrado de un número aumentado en 1. ⦁ Cinco veces un número menos 8. ⦁ El número siguiente a un múltiplo de 5. ⦁ Un número impar.

Answers

Answer:

1) N^2 + 1

2) 5*N - 9

3) n*5 + 1    (con n entero)

4) n*2 + 1    (con n entero)

Step-by-step explanation:

1)  El cuadrado de un número aumentado en 1.

Podemos definir a N como el número.

El cuadrado de N es:

N^2

Y a esto lo tenemos que aumentar en 1, entonces tenemos:

N^2 + 1

Esto es el cuadrado de un número aumentado en 1.

2)  Cinco veces un número menos 8.

Esto es:

5*N  (5 veces un número)

Y a esto le restamos 8, para obtener:

5*N - 8

3) El número siguiente a un múltiplo de 5

Un multiplo de 5 se escribe como:

n*5

Donde n es un número entero.

El numero siguiente es:

n*5 + 1

4) Un número impar.

Bien, sabemos que un número par es un multiplo de 2.

Entonces:

n*2 es un número par.

Entonces, n*2 + 1 tiene que ser un numero impar.

En este caso la expresión algebraica es:

n*2 + 1
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