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Nitella [24]
3 years ago
11

Factor

xy^2+32y^3" align="absmiddle" class="latex-formula"> completely.
Mathematics
2 answers:
Darya [45]3 years ago
7 0

Answer:

y(3x + 4y)(x + 8y)

General Formulas and Concepts:

<u>Alg I</u>

  • Factoring

Step-by-step explanation:

<u>Step 1: Define expression</u>

3x²y + 28xy² + 32y³

<u>Step 2: Simplify</u>

  1. Factor out <em>y</em>:                    y(3x² + 28xy + 32y²)
  2. Factor trinomial:              y(3x + 4y)(x + 8y)
  • What multiplies up to 32 adds up to 28 with 3 being multiplied as a coefficient? Numbers are 4 and 8.
balandron [24]3 years ago
6 0

SOLVING STEPS...

<u>STEP</u><u>1</u><u> </u>

FACTOR OUT y FROM THE EXPRESSION

SO ITS

y \times (3x {}^{2}  + 28xy + 32y {}^{2} )

<u>STEP</u><u>2</u><u> </u>

<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>WRITE 28y AS A SUM

<u>STEP3</u>

<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>FACTOR OUT 3x2 + 24xy

SO IT'S

y \times (3x \times (x + 8y) + 4xy + 32y {}^{2} )

<u>STEP</u><u>4</u>

<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>FACTOR OUT 4xy + 32y2

SO IT'S

y \times (3x \times (x + 8y) + 4y \times (x + 8y) \:   )

<u>STEP5</u><u> </u>

FACTOR OUT (3x × ( x + 8y ) + 4y × ( x + 8y ) )

SO THE ANSWER IS

y \times (x + 8y) \times (3x + 4y)

GAVE ME THE BRAINLIEST IF MY ANSWER WAS HELPFUL...

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Which of the following statements about the polynomial function f(x)=x^3+2x^2-1
ch4aika [34]

x = -1

x =(1-√5)/-2= 0.618

x =(1+√5)/-2=-1.618

Step  1  :

Equation at the end of step  1  :

 0 -  (((x3) +  2x2) -  1)  = 0  

Step  2  :  

Step  3  :

Pulling out like terms :

3.1     Pull out like factors :

  -x3 - 2x2 + 1  =   -1 • (x3 + 2x2 - 1)  

3.2    Find roots (zeroes) of :       F(x) = x3 + 2x2 - 1

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  x  which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient

In this case, the Leading Coefficient is  1  and the Trailing Constant is  -1.

The factor(s) are:

of the Leading Coefficient :  1

of the Trailing Constant :  1

Let us test ....

  P    Q    P/Q    F(P/Q)     Divisor

     -1       1        -1.00        0.00      x + 1  

     1       1        1.00        2.00      

The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms

In our case this means that

  x3 + 2x2 - 1  

can be divided with  x + 1  

Polynomial Long Division :

3.3    Polynomial Long Division

Dividing :  x3 + 2x2 - 1  

                             ("Dividend")

By         :    x + 1    ("Divisor")

dividend     x3  +  2x2      -  1  

- divisor  * x2     x3  +  x2          

remainder         x2      -  1  

- divisor  * x1         x2  +  x      

remainder          -  x  -  1  

- divisor  * -x0          -  x  -  1  

remainder                0

Quotient :  x2+x-1  Remainder:  0  

Trying to factor by splitting the middle term

3.4     Factoring  x2+x-1  

The first term is,  x2  its coefficient is  1 .

The middle term is,  +x  its coefficient is  1 .

The last term, "the constant", is  -1  

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

Step-2 : Find two factors of  -1  whose sum equals the coefficient of the middle term, which is   1 .

     -1    +    1    =    0  

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Equation at the end of step  3  :

 (-x2 - x + 1) • (x + 1)  = 0  

Step  4  :

Theory - Roots of a product :

4.1    A product of several terms equals zero.  

When a product of two or more terms equals zero, then at least one of the terms must be zero.  

We shall now solve each term = 0 separately  

In other words, we are going to solve as many equations as there are terms in the product  

Any solution of term = 0 solves product = 0 as well.

Parabola, Finding the Vertex :

4.2      Find the Vertex of   y = -x2-x+1

For any parabola,Ax2+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is  -0.5000  

Plugging into the parabola formula  -0.5000  for  x  we can calculate the  y -coordinate :  

 y = -1.0 * -0.50 * -0.50 - 1.0 * -0.50 + 1.0

or   y = 1.250

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = -x2-x+1

Axis of Symmetry (dashed)  {x}={-0.50}  

Vertex at  {x,y} = {-0.50, 1.25}  

x -Intercepts (Roots) :

Root 1 at  {x,y} = { 0.62, 0.00}  

Root 2 at  {x,y} = {-1.62, 0.00}  

Solve Quadratic Equation by Completing The Square

4.3     Solving   -x2-x+1 = 0 by Completing The Square .

Multiply both sides of the equation by  (-1)  to obtain positive coefficient for the first term:

x2+x-1 = 0  Add  1  to both side of the equation :

  x2+x = 1

Now the clever bit: Take the coefficient of  x , which is  1 , divide by two, giving  1/2 , and finally square it giving  1/4  

Add  1/4  to both sides of the equation :

 On the right hand side we have :

  1  +  1/4    or,  (1/1)+(1/4)  

 The common denominator of the two fractions is  4   Adding  (4/4)+(1/4)  gives  5/4  

 So adding to both sides we finally get :

  x2+x+(1/4) = 5/4

Adding  1/4  has completed the left hand side into a perfect square :

  x2+x+(1/4)  =

  (x+(1/2)) • (x+(1/2))  =

 (x+(1/2))2

Things which are equal to the same thing are also equal to one another. Since

  x2+x+(1/4) = 5/4 and

  x2+x+(1/4) = (x+(1/2))2

then, according to the law of transitivity,

  (x+(1/2))2 = 5/4

We'll refer to this Equation as  Eq. #4.3.1  

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of

  (x+(1/2))2   is

  (x+(1/2))2/2 =

 (x+(1/2))1 =

  x+(1/2)

Now, applying the Square Root Principle to  Eq. #4.3.1  we get:

  x+(1/2) = √ 5/4

Subtract  1/2  from both sides to obtain:

  x = -1/2 + √ 5/4

Since a square root has two values, one positive and the other negative

  x2 + x - 1 = 0

  has two solutions:

 x = -1/2 + √ 5/4

  or

 x = -1/2 - √ 5/4

Note that  √ 5/4 can be written as

 √ 5  / √ 4   which is √ 5  / 2

Solve Quadratic Equation using the Quadratic Formula

4.4     Solving    -x2-x+1 = 0 by the Quadratic Formula .

According to the Quadratic Formula,  x  , the solution for   Ax2+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :

                                     

           - B  ±  √ B2-4AC

 x =   ————————

                     2A

 In our case,  A   =     -1

                     B   =    -1

                     C   =   1

Accordingly,  B2  -  4AC   =

                    1 - (-4) =

                    5

Applying the quadratic formula :

              1 ± √ 5

  x  =    ————

                  -2

 √ 5   , rounded to 4 decimal digits, is   2.2361

So now we are looking at:

          x  =  ( 1 ±  2.236 ) / -2

Two real solutions:

x =(1+√5)/-2=-1.618

or:

x =(1-√5)/-2= 0.618

Solving a Single Variable Equation :

4.5      Solve  :    x+1 = 0  

Subtract  1  from both sides of the equation :  

                     x = -1

Hope this helps.

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THIS IS THE COMPLETE QUESTION

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No, the shot is not blocked.

Answer:

Yes, between 0.84 seconds and 0.85 seconds after the shot is launched.

Given that the

equation for shot block height h = 9 + 25t – 16t2.

equation for ball height h = 6 + 30t – 16t2.

From the question ,it can be deducted that the shot is made before two tenths of a second or 0.2seconds

CHECK THE ATTACHMENT TO COMPLETE THE DETAILED SOLUTION

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Given

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