Factor completely & explain please...?

Question

v^5+4v^4f-60v^3f^2

and

16-w^4p^4

Answer 1

Hi,

v^5+4v^4f-60v^3f^2

Factor out the GCF of v³.

v³(v² + 4vf - 60f²)

Factor the trinomial.

v³(v + 10f)(v -6f) <== answer

and

16-w^4p^4

This is the difference of perfect squares.

(4 - w²p²)(4 + w²p²)

The first factor is also the difference of perfect squares. It factors again.

(2 - wp)(2 - wp)(4 + w²p²) <== answer


I hope that helps!! :-)

Answer 2

For the 1st question:
v^5 + 4v^4f - 60v³f²

You can simplify this by taking on the variable v first: v³ (v² + 4vf - 60f²)

Now, you can further simplify the equation inside the parenthesis: (v + 10f) (v - 6f)

Put them all together the answer is: v³ (v + 10f) (v - 6f)


For the second question:
16 - w^4p^

This is almost a perfect square if it was the + sign. Neverthe less, the factors are:

(4 + w²p²) (4 - w²p²)

But take note, (4 - w²p²) can still be simplified. The final answer is:

(4 + w²p²) (2 - wp) (2 + wp)

Answer 3

In the first expression, v^2 is common to each term:

v^3(v^2 + 4vf - 60f^2)

The second part can be factored as (v + 10f)(v - 6f). To do this, you search for two numbers that multiply to -60f^2 (the last coefficient) and add to 4f (the middle coefficient). These two numbers are 10f and -6. So you get (v + 10f)(v - 64). The complete answer is:

v^3(v + 10f)(v - 6f)

To do the second one, you must notice that the expression is a difference of squares, meaning it is of the form (x^2 - y^2), where x = 4, and y = (wp)^2. Anytime you have a difference of squares (x^2 - y^2), it can be factored as (x + y)(x - y). Therefore, you have:

(4 + (wp)^2)(4 - (wp)^2)

Answer 4

v^5 + 4v^4f - 60v^3f^2
v^3(v^2 + 4vf - 60f^2) {common factor}

Look for factors of - 60 whose sum is 4

-6, 10 do it

v^3(v - 6f)(v + 10f)

--------------------

16-w^4p^4

Difference of two squares a^2 - b^2 = (a + b)(a - b)

(4 + w^2p^2)(4 - w^2p^2)
(4 + w^2p^2)(2 + wp)(2 + wp)

Answer 5

In the first expression take out a common factor of v^3 and you get v^3(v^2+4vf-60f^2)

use foil to factor the quadratic and final answer is

v^3(v+10f)(v-6f)

second expression facor using foil into
(4-w^2p^2)(4+w^2p^2) now the first term is factorable so you end up with
(2+wp)(2-wp)(4+w^2p^2)

Answer 6

Question 1
v ³ (v ² + 4 v f - 60 f ²)
v ³ (v + 10f) (v - 6f)

Question 2
(4 - w ² p ²) (4 + w ² p ²)
(2 - w p) (2 + w p) (4 + w ² p ²)