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MAXImum [283]
3 years ago
7

Factor out the GCF from the terms of the polynomial –4y5 + 6y3 + 8y2 – 2y. A. –2y4 + 3y2 + 4y – 1 B. y(–4y4 + 6y2 + 8y – 2) C. 2

y(–2y4 + 3y2 + 2y) D. –2y(2y4 – 3y2 – 4y + 1)
Mathematics
1 answer:
andreyandreev [35.5K]3 years ago
8 0

Answer:

Option D) -2y(2y^4-3y^2-4y+1) is correct

Therefore -4y^5+6y^3+8y^2-2y=-2y(2y^4-3y^2-4y+1)

Step-by-step explanation:

Given polynomial is -4y^5+6y^3+8y^2-2y

To factorise the given polynomial by taking out the common terms of the given polynomial :

  • -4y^5+6y^3+8y^2-2y
  • =2y(-2y^4+3y^2+4y-1) ( here 2y is GCF common term so taking outside the terms of the polynomial )
  • =-2y(2y^4-3y^2-4y+1) ( now taking (-) outside )

Therefore -4y^5+6y^3+8y^2-2y=-2y(2y^4-3y^2-4y+1)

Option D) -2y(2y^4-3y^2-4y+1) is correct

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5-2x2=-15

Two solutions were found :

                  x = ± √10 = ± 3.1623

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                    5-2*x^2-(-15)=0

Step by step solution :

Step  1  :

Equation at the end of step  1  :

 (5 -  2x2) -  -15  = 0

Step  2  :

Step  3  :

Pulling out like terms :

3.1     Pull out like factors :

  20 - 2x2  =   -2 • (x2 - 10)

Trying to factor as a Difference of Squares :

3.2      Factoring:  x2 - 10

Theory : A difference of two perfect squares,  A2 - B2  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =

        A2 - AB + BA - B2 =

        A2 - AB + AB - B2 =

        A2 - B2

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check : 10 is not a square !!

Ruling : Binomial can not be factored as the difference of two perfect squares.

Equation at the end of step  3  :

 -2 • (x2 - 10)  = 0

Step  4  :

Equations which are never true :

4.1      Solve :    -2   =  0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation :

4.2      Solve  :    x2-10 = 0

Add  10  to both sides of the equation :

                     x2 = 10

When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  

                     x  =  ± √ 10  

The equation has two real solutions  

These solutions are  x = ± √10 = ± 3.1623  

Two solutions were found :

                  x = ± √10 = ± 3.1623

6 0
2 years ago
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