Can someone show me how to factor these 2 equations?

Question

a. 4x^3-8x^2-25x+50
b. x^4-1

Answer 1

a) 4x^3 - 8x^2 - 25x + 50
= 4x^2(x - 2) - 25(x - 2)
= (x - 2)(4x^2 - 25)
= (x - 2)[(2x)^2 - 5^2]
= (x - 2)(2x + 5)(2x - 5)

b) x^4-1
= (x^2)^2-1^2
= (x^2 + 1)(x^2 - 1)
= (x^2 + 1)(x + 1)(x - 1)

Answer 2

These two equations deal with differences of perfect squares.
A square is a term that is derived from multiplying something by itself, like 4 (2*2), x^2 (x*x), or 4x^2 (2x*2x)
When a perfect square is subtracted from another perfect square, you can use a property: a^2 - b^2 = (a-b)(a+b)


a.) Group the equation like this: (4x^3-8x^2)+(-25x+50)
Then factor out: 4x^2(x-2) - 25(x-2)
Now you have: (4x^2 - 25)(x-2)
The first half is a perfect square: (2x-5)(2x+5)(x-2)
b.) Another perfect square: (x^2-1)(x^2+1)

Answer 3

a. 4x^3-8x^2-25x+50
= ( 4 x^3 - 8 x^2 ) + ( -25x + 50 )
= 4x^2 ( x - 2 ) - 25 ( x - 2 )
= ( x - 2 ) ( 4x^2 - 25 )
= ( x - 2 ) ( 2x - 5 ) ( 2x + 5 )

b. x^4-1 = ( x^2 - 1 ) ( x^2 + 1 ) = (x - 1 ) ( x + 1 ) ( x^2 + 1 )

NOTICE
these are called algebric expression

Answer 4

a) Factor by grouping. 4x^3-8x^2-25x+50 = 4x^2(x-2)-25(x-2) = (x-2)(4x^2-25) = (x-2)(2x+5)(2x-5)

b) Difference of squares: x^4-1 = (x^2+1)(x^2-1) = (x^2+1)(x+1)(x-1)

Answer 5

Factoring is crap. lol For a. you pretty much just have to divide by random numbers (x-1, x+3, x-4, etc.) and just see which one comes out without any remainders, and that means that it is a factor. Then you take whatever you got after you divided, and do the same thing until you can't do it anymore.
For b... I don't know how to explain it, but the answer is...
(x^2+1)(x^2-1) then you factor that to...
(x+i)(x-i)(x+1)(x-1).. where i is the square root of -1.

Answer 6

You have to take out the greatest common factor that goes into all terms in an equation
a) 4x^2(x-2)-25(x-2)
(4x^2-25)(x-2) <--- if you use FOIL, it should factor out
b) x^4-1
(x^2-1)(x^2+1)
(x-1)(x+1)(x^2+1)

Answer 7

a.
4x^3 - 8x^2 - 25x + 50

4x^2(x - 2) - 25(x - 2 )

(4x^2 - 25)(x - 2)

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

b.
x^4 -1

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

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

Answer 8

Ques: x^4 -1 ans=> this can be written as (x^2)^2 - 1
But we have the formula (a^2 - b^2) = (a+b)(a-b) => {(x^2)+1}{(x^2)-1} again this can be factored into {(x+1)(x-1)}{(x+1)(x-1)}
Hence x^4 - 1 = (x+1)(x-1)(x+1)(x-1)

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Answer 9

1st Subtract the 360 from the two factors (-10y^2=-360) 2d Divide via -10 from the two facet (y^2=-360/10) third destructive 360 divided via destructive 10 equals 36. (y^2=36) 4th Take the sq. Root of the two factors now. y=6

Answer 10

a. 4x^2(x-2) - 25(x-2)
(x-2)(4x^2-25)
(x-2)(2x -5)(2x+5)

b. (x^2-1)(x^2+1)
(x+1)(x-1)(x^2+1)