Factor w^4=17w^2-16...please help?

Question

also solve inequality..x^2-5x>6

i need practice for a placement test tomorrow and i dont understand thins thxs

Answer 1

So you want to factor

w^4 = 17w^2 - 16

First, move everything over to the left hand side.

w^4 - 17w^2 + 16 = 0

Now, factor. This is a quadratic in disguise; you just may not realize it yet.

(w^2 - 16)(w^2 - 1) = 0

Both of these are difference of squares, so you have to factor them as well. Note that a^2 - b^2, a difference of squares, factors into (a - b) (a + b).

(w - 4)(w + 4) (w - 1)(w+1) = 0

Therefore, w = 4, -4, 1, -1

To solve the inequality x^2 - 5x > 6, first, do the following.

Move 6 to the left hand side.

x^2 - 5x - 6 > 0

Now, factor the left hand side.

(x - 6) (x + 1) > 0

This means that x = 6 and x = -1 are critical points. What you have to do at this point is to test the behaviour AROUND those points.

You want to test three intervals: (-infinity, -1), (-1, 6), and (6, infinity).

Test a point in the first interval: -100. Do you get a true or false result, when plugging it into (x - 6) (x + 1) > 0? The answer is true, because you get a negative times a negative, which is a positive, and positive numbers are greater than 0. Therefore, we include that interval.

Test a point in the second interval: 0. Plug in 0 for (x - 6) (x + 1) > 0 and we get (-6)(1), which is a negative number. Therefore, we exclude that interval.

Test a point in the third interval: 100. We get a positive number, so we include that interval.

Therefore, our results showed that

(-infinity, -1) passed our test.
(-1, 6) failed our test.
(6, infinity) passed our test

So x = (-infinity, -1) U (6, infinity)

Answer 2

Rearrange to form

w^4 - 17w^2 - 16 = 0 let z = w^2

z^2 - 17z - 16 = 0

(z-1)(z-16) = 0 put back z = w^2

(w^2 - 1)(w^2 - 16) Each of these is a difference of squares that factor into the form (a+b)(a-b) = a^2 - b^2. The factors are then

(w+1)(w-1)(w+4)(w-4)


For the second, put all terms on the left

x^2 - 5x - 6 > 0

This factors to (x + 1)(x - 6) > 0

For the product of two terms > 0, both terms are > 0 or both terms are < 0; Both terms are > 0 only if x > 6; both are < 0 only if x < -1

Answer 3

1} w^4=17w^2-16
=w^4+w^2=17-16
=w^6=1
=w^6

2} x^2-5x>6
x^2-5x>6
5x-x^2=6
4x^2=6
4x^=6-2
4x^=4
x^=4/4
x^=1

Answer 4

w^4 - 17w² + 16 = 0
(w² -1) (w² - 16) = 0
(w -1) (w + 1) (w + 4) (w - 4) = 0

Answer 5

(w^2 -1) (w^2 - 16)

Answer 6

w^4=17w^2-16
w^4 - w^2=16w^2-16
w^2(w^2-1)=16(w^2-1)
w^2=16
w=4