How do you factor polynomials?

Question

2x^2-50
8x^6-32x^5+4x^4

Answer 1

Let's see,

first equation... 2x^2 - 50

notice that 2 is common to both 2x^2 and 50.

2(x^2-25)

and noticing that you have a a^2 - b^2 condition here,

2(x-5)(x+5)........that's the result for 1

For the second polynomial,

4 is a common factor and so is x^4

4x^4(2x^2-8x+1)

Unfortunately, the roots of the quadratic expression contain the square roots of negative numbers.

Good Luck.

Answer 2

2x^2 - 50.
First reduce as much as possible, leaving us with 2(x^2 - 25). Now we need to factor the term x^2 - 25. Since the term is a difference of squares, we know we'll have two factors multiplying the square roots of the term, meaning (x + 5) times (x - 5). (The reason *why* we know there'll be two factors is rather complicated, and you should find a tutor to demonstrate this). So, the final answer is 2(x + 5)(x - 5).

You sure you wrote the second polynomial correctly? That one can't be factored, beyond reducing it to (4x^4)(2x^2 - 8x + 1).

Answer 3

2x^2 - 50 = 2(x^2 - 25) = 2(x - 5)(x + 5)
8x^6 - 32x^5 + 4x^4 = 4x^4(2x^2 - 8x + 1)

Answer 4

1) 2(x+5)(x-5)

2) This one must have a typo, because the most you can factor it is 4x^4(2x^2-8x+1). In order to end up with a "+!) at the end, both polynomials would have to end in "-1", making the "-8x" impossible in my book.....

Hope this helped!

Answer 5

is there an equals sign somewhere?
or are those 2 separate equations?
2(x^2-25) also 2(x-5)(x+5)
4x^4(2x^2-8x+1)

Answer 6

quadtratic equation

ax^2 + bx + c = 0

x1 = (-b + sqrt(b^2 - 4ac)) / 2a
x2 = (-b - sqrt(b^2 - 4ac)) / 2a

factored form is
a * (x - x1)*(x - x2)

1) a = 2, b = 0, c = -50
x1 = sqrt(400)/4 = 5
x2 = -sqrt(400)/4 = -5

2*(x-5)(x+5)

2) (8x^2 - 32x + 4) * x^4

a = 8, b = -32, c = 4, find x1,x2 using equation,
then x3 = x4 = x5 = x6 = 0

x1 = (32 + sqrt(1024 - 128)) / 16 = 3.87
x2 = (32 - sqrt(1024 - 128)) / 16 = 0.129

(8x^4)(x-3.87)(x-0.129)

Answer 7

are these two separate problems? if so...

the first one (2x^2-50) would break down into 2(x-5)(x+5).

the second one would be imaginary, because no real solution works... you would have to use the quadratic formula for that

Answer 8

2x^2 - 50
= 2x^2 - 50
= 2 (x^2 - 25)
= 2 (x - 5) (x + 5)

8x^6 - 32x5 + 4x^4
= 8x^6 - 32x5 + 4x^4
= 8x^4 (x^2 - 4x + 0.5)
= 8x^4 (x - 3.87) (x - 1.13)

Answer 9

Next

Answer 10

2x^2-50=
2(x^2-25)=
2(x-5)(x+5)