Need help factoring this binomial: x^4-1?

2 answers:

8 0

So we notice that is the special factorization known as 'the difference of two perfect squares'
exg
a^2-b^2=(a-b)(a+b) so
x^4=(x^2)^2 and 1=1^2 so
(x^2)^2-(1)^2=(x^2-1)(x^2+1)

natima [27]4 years ago

5 0

Use the difference of squares factorization - that for any numbers a and b, (a-b)(a+b)=a^2-b^2.

We have:

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

In addition:

(x-1)(x+1)=x^2-1, so we have:

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

As our complete factorization.

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Viefleur [7K]

40 gold

5 gold:3 red = x gold:24 red
24 × 5= 120
120/3 = 40

3 0

4 years ago

at is another way to write the expression p⋅(10−2) ? (10⋅2)−p 10⋅2−p (10⋅2)⋅p (p⋅10)−(p⋅2)

Ostrovityanka [42]

Answer:

(p ⋅ 10) − (p ⋅ 2)

Step-by-step explanation:

Try to substitute "p" with any number.

Let's say p = 2

p ⋅ (10 − 2)

2 ⋅ (10 − 2)

2 ⋅ (8)

= 16

Now for the other equations

(p ⋅ 10) − (p ⋅ 2)

(2 ⋅ 10) − (2 ⋅ 2)

(20) - (4)

= 16

(10 ⋅ 2) − p

(10 ⋅ 2) − 2

(20 ) − 2

= 18

10 ⋅ 2 − p

10 ⋅ 2 − 2

20 - 2

= 18

(10 ⋅ 2) ⋅ p

(10 ⋅ 2) ⋅ 2

(20) ⋅ 2

= 40

Only (p ⋅ 10) − (p ⋅ 2) has the same result as p ⋅ (10 − 2).

Cheers

7 0

3 years ago

I need help on two questions!

Inessa05 [86]

1) Answer: $\frac{x^{2}}{100} +\frac{y^{2}}{4} = 1$

<u>Explanation:</u>

The equation of an ellipse is: $\frac{(x-h)^{2}}{a^{2}} +\frac{(y-k)^{2}}{b^{2}} = 1$ ; where (h, k) is the center, "a" is the x-radius, and "b" is the y-radius.