help with factoring equations?

1) Let y = x^3 then y^2 = x^6, using this the equation becomes:

y^2 - 9y + 8 = 0

y^2 - 8y - y + 8 = 0

y (y - 8) - 1 (y - 8) = 0

(y - 1)(y - 8 ) = 0

y = 1 OR y = 8

Now re-call that x^3 = y so:

x^3 = 1

x = ³√1 = 1 ANSWER.

AND

x^3 = 8

x = ³√8

x = 2 ANSWER.
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2) 16x² - 64 = 0

16x² - 8² = 0

(4x)² - (8)² = 0

(4x - 8)(4x + 8) = 0

x = 8/4 = 2
OR x = -8/4 = -2 ANSWERS.

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3) x² + 5x + 4 = 0

x² + 4x + x + 4 = 0

x (x + 4) + 1 (x + 4) = 0

(x + 1) (x + 4) = 0

x = -1 OR x = -4 ANSWER.
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4) 2x² + 3x - 2 = 0 <<<< ASSUMING that this is your question

2x² + 4x - x - 2 = 0

2x (x + 2) - 1 (x + 2) = 0

(2x - 1) (x + 2) = 0

x = 1/2 OR x = -2 ANSWER.

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5) x² - 6x = 0

x (x - 6 ) = 0

x = 0 ANSWER.

OR x - 6 = 0 ==> x = 6 ANSWER.

1) x^6 - 9x^3 +8=0
= (x^3 - 1)(x^3 - 8)
= (x - 1)(x^2 + x + 1)(x - 2)(x^2 + 2x + 4) = 0
(x - 1) = 0
x = 1
(x^2 + x + 1) = 0
x = {-1 ± √(1 - 4)}/2
x = {-1 ± i√3)/2
(x - 2) = 0
x = 2
(x^2 + 2x + 4) = 0
x = {-2 ± √(4 - 16)}/2
x = {-2 ± √- 12}/2
x = {-2 ± 2√-3}/2
x = -1 ± i√3

2)16x^2 - 64=0
= 16(x^2 - 4)
= 16(x + 2)(x - 2)

3) x^2 + 5x +4=0
(x + 1)(x + 4) = 0
(x + 1) = 0
x = -1
(x + 4) = 0
x = -4

4) 2x^2 = 3x -2=0
2x^2 - 3x - 2 = 0
(2x + 1)(x - 2) = 0
(2x + 1) = 0
2x = -1
x = -1/2
(x - 2) = 0
x = 2

5) x^2 -6x=0
x(x - 6) = 0
x = 0
(x - 6) = 0
x = 6

1) Normally, anything with a higher power than x² is difficult if not impossible to factorize, so if you get a much higher power, as in this instance, either it cannot be done, or there is a "trick", you might say.

Here you have to recognize that x^6 and x^3 can both be represented as powers of x^3 - (x^3)² and (x^3)¹. So if you substitute for x^3, you get just a quadratic expression, which you can factorize.

Let x^3 = X, say,

then X² - 9X + 8 = 0

(If you are just learning factorization, you always start with the two sets of brackets : (X ± ... ) (X ± ...) Of course if you have 3X² or something then you must include the 3 in one of the factors.

Now look at the constant term, 8. You want two numbers which give +8 when multiplied, but give -9 when added. You could have ±2 and ±4, but you cannot get a -9 from these. The other possibility here is ±1 and ±8, which will give you either +6 or -7 if one of them is + and the other -, or +9 or -9 if both are + or -. The option you want here is obviously -1 and -8, so you can fill in the blanks in the brackets :

(X - 1) (X - 8) (that looks long-winded perhaps, but it does get easier with practise).

Now you can replace the X with the x^3's, giving you finally

x^6 - 9x^3 + 8 = (x^3 - 1) (x^3 - 8)

(I note that it is the factorization which is bugging you, but if you want to proceed to a solution with this one, you simply proceed as normal for a quadratic :

x^3 = 1 or x^3 = 8, which gives you x = 1 or x = 2 as two solutions.

A polynomial has as many solutions as there are degrees in the highest power, however, so there are in all 6 possible solutions to this equation. Usually most, if not all, are not real numbers, however, as in this case. But you will come to the others soon enough, you have enough to worry about for now !
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2) 16x² - 64 : It is always wise to start by looking just at the numbers, to see if there is a common numerical factor - this often simplifies the factorization of the rest. Here, you can start by taking out a factor of 16, giving you 16(x² - 4)

Then you should recognize that the bracketed part is what is called "the difference of two squares" (it crops up over and over again, and you do need to memorize it) : (a² - b²) = (a + b) (a - b).

So the whole thing becomes 16(x + 2) (x - 2)
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3) This is a relatively straightforward one :

x² + 5x + 4 = (x + ...) (x + ...)

The possible factors of 4 are 2, 2, and 1, 4 of course, and it is obvious that 1 and 4 are the ones required.

So the answer is (x + 1) (x + 4)
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4) I presume that this is meant to be 2x² + 3x - 2 = 0

Here you have to include the 2 of 2x² in one of your brackets :

2x² + 3x - 2 = (2x ± ... ) (x ± ... )

The only possible factors for the constant term, -2, are ±1 and ±2 (but this time it is important which way round you put them into the brackets, because the 2 in 2x makes a big difference to the result.)

The only way to get the correct placement is by trial and error.
(2x + 2) (x - 1) gives you 2x² - 2x + 2x - 2 = 2x² - 2, which is clearly wrong. So the right placement is

(2x - 1) (x + 2) which gives 2x² + 3x - 2 as required.
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5) x² - 6x - this is another slightly "tricky" one for beginners, because you are expecting (you have been primed to expect) a standard quadratic, and it is off-putting to find one which is lacking a constant term. However, this makes it, in fact, slightly easier to deal with, since there is the factor x to be extracted, which gives you

x² - 6x = x (x - 6) = 0

and since one of the factors must be 0, then you can see immediately that one possible solution is x = 0. The other solution, of course comes from (x - 6) = 0, hence x = 6
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1)
x^6 - 9x^3 + 8 = 0
x^6 - x^3 - 8x^3 + 8 = 0
(x^6 - x^3) - (8x^3 - 8) = 0
x^3(x^3 - 1) - 8(x^3 - 1) = 0
(x^3 - 1)(x^3 - 8) = 0
(x - 1)(x^2 + x + 1)(x - 2)(x^2 + 2x + 4) = 0

x - 1 = 0
x = 1

x^2 + x + 1 = 0
x = [-b ±√(b^2 - 4ac)]/2a
a = 1
b = 1
c = 1
x = [-1 ±√(1 - 4)/2
x = [-1 ±√(-3)]/2 (imaginary number)

x - 2 = 0
x = 2

x^2 + 2x + 4 = 0
x = [-b ±√(b^2 - 4ac)]/2a
a = 1
b = 3
c = 4
x = [-3 ±√(9 - 16)]/2
x = [-3 ±√(-7)]/2 (imaginary number)

∴ x = 1 , 2 , [-1 ±√(-3)]/2 , [-3 ±√(-7)]/2

= = = = = = = =

2)
16x^2 - 64 = 0
(16x^2 - 64)/16 = 0/16
x^2 - 4 = 0
x^2 + 2x - 2x - 4 = 0
(x^2 + 2x) - (2x + 4) = 0
x(x + 2) - 2(x + 2) = 0
(x + 2)(x - 2) = 0

x + 2 = 0
x = -2

x - 2 = 0
x = 2

∴ x = -2 , 2

= = = = = = = =

3)
x^2 + 5x + 4 = 0
x^2 + 4x + x + 4 = 0
(x^2 + 4x) + (x + 4) = 0
x(x + 4) + 1(x + 4) = 0
(x + 4)(x + 1) = 0

x + 4 = 0
x = -4

x + 1 = 0
x = -1

∴ x = -4 , -1

= = = = = = = =

4)
2x^2 - 3x - 2 = 0
2x^2 + x - 4x - 2 = 0
(2x^2 + x) - (4x + 2) = 0
x(2x + 1) - 2(2x + 1) = 0
(2x + 1)(x - 2) = 0

2x + 1 = 0
2x = -1
x = -1/2 (-0.5)

x - 2 = 0
x = 2

∴ x = -1/2 (-0.5) , 2

= = = = = = = =

5)
x^2 - 6x = 0
x(x - 6) = 0

x = 0

x - 6 = 0
x = 6

∴ x = 0 , 6

1)(x-2)(x^2+2x+4)(x-1)(x^2+x+1)=0
2)16(x-2)(x+2)=0
3)(x+1)(x+4)=0
4)(2x-1)(x+2)=0
5)x(x-6)=0

1. (x^3 - 8) (x^3 -1) = 0

2. (4x + 8) (4x - 8) = 0

3. (x + 4) (x + 1) = 0

4. do you mean 2x^2 + 3x -2 ???

(2x-1 ) (x + 2) = 0

5. x ( x- 6) = 0


you just have to be familiar with it, then you can solve this just by inspection.

good luck.

I believe you asked to factor the equations, not solve, so here are the factored equations. (all equal to 0)

1. (x^3 - 8)(x^3 - 1)

2. 16(x^2 - 4)

3. (x + 4)(x + 1)

4. -> 2x^2 - 3x + 2 --> (2x + 1)(x - 2)

5. x(x - 6)

let x^3 = y
1/ y^2 - 9y + 8 = 0----> y = 1,8
x^3 = 1 --->x=1
x^3 = 8--->x= 2

2)16x^2 - 64=0
16(x^2-4) = 0
16(x-2)(x+2) = 0--->x = 2, -2