DIFFICULT MATH QUESTION!! HELP NEEDED PLEASE!!?

If (x-k) is a factor of this expression, then, k is a root of the equation: kx^3 + 5x^2 -7kx -8 = 0. You replace x by k, and you get: k^4 - 2k^2 -8 = 0 (simple calculation), then, you take Y=k^2, you replace Y in the preceding equation, so you get: Y^2 -2Y -8 =0, which is a typical quadratic equation, you solve it with a condition that Y is greater or equal to 0, and get that it has two roots: Y=-2 (rejected) and Y=4 (accepted) (since Y is greater or equal to 0). Hence, we have to values of k: -2 and 2 (the square roots of 4). But, k is a positive integer, hence, k=-2 is rejected and the only value of k is: k = 2 :)

Hence the above expression becomes: 2x^3 + 5x^2 -14x - 8

You get the other factor(s) by using the following definition: If (x-k) is a factor of this expression, then, this expression can be written in the form: (x-k)(ax^2 +bx +c) Where a, b & c are all real constants. You already know that k is equal to 2, then the expression can be written as: (x-2)(ax^2 +bx +c). After you expand the preceding product, you will get: ax^3 + (b - 2a)x^2 + (c - 2b)x -2c.

Then, you compare the obtained result with the expression above, as follows:

2x^3 + 5x^2 - 14x - 8 = ax^3 + (b - 2a)x^2 + (c - 2b)x - 2c

So: (1) a=2

(2) b - 2a = 5, so b - 4 = 5, then b=9

(3) -2c = -8, then c=4

Therefore the expression can be written as: (x - 2)(2x^2 - 9x +4)

You can as well factorize (2x^2 -9x +4) which is in the form: (ax^2 +bx +c) by determining its roots which are: x=1/2 and x=4. Eventually you will obtain that 2x^2 -9x +4 = (2x-1)(x-4)

Eventually, the expression can be finally written as: (x-2)(x-4)(2x-1)

I know I have mentioned a lot of details but that is just to make the idea as clear as possible. I hope this was enough :)

The factor theorem states that x-k is a factor of a polynomial if and only the polynomial evaluates to zero at x=k. Therefore, we have

k(k^3) + 5k^2 - 7k(k) - 8 = 0

k^4 - 2k^2 - 8 = 0

(k^2)^2 - 2k^2 - 8 = 0

y^2 - 2y - 8 = 0, letting y = k^2

(y + 2)(y - 4) = 0

y = -2 or y = 4

k^2 = -2 or k^2 = 4

k = 2 is the only positive solution for k.

So we need to factor 2x^3 + 5x^2 - 14x - 8, and we already know that x - 2 is a factor. Divide (2x^3 + 5x^2 - 14x - 8) by (x - 2) using either synthetic division:

2 on the outside

2 5 -14 -8

_ 4 18 8

2 9 4 0

The quotient is 2x^2 + 9x + 4, and as expected, the remainder is zero. Finally, we have

2x^3 + 5x^2 - 14x - 8 = (x - 2)(2x^2 + 9x + 4)

= (x - 2)(2x^2 + 8x + x + 4), from seeing that 8 and 1 have sum 9 and product 2*4 = 8

= (x - 2)[2x(x + 4) + 1(x + 4)], from beginning to factor the trinomial by grouping

= (x - 2)(x + 4)(2x + 1), from finishing the factoring by grouping.

Lord bless you today!

(x - k) is a factor of a polynomial P(x) if P(k) = 0

P(x) = kx^3 + 5x^2 - 7kx - 8

P(k) = k(+k)^3 + 5(+k)^2 - 7k(+k) - 8 = k^4 - 2k^2 - 8

P(k) = 0 if k^4 - 2k^2 - 8 = 0

set k^2 = w, so that k^4 = (k^2)^2 = w^2

The equation becomes

w^2 - 2w - 8 = 0

w = - 2, w = 4

substitute back k^2

k^2 = -2 has no real solutions

k^2 = 4

k = ±2

if k = 2

P(x) = 2x^3 + 5x^2 - 14x - 8

synthetic division

__________2______5_______-14_____|__-8

______2__________4________18_____|__8

__________2______9________4______|__0

Quotient Q(x) = 2x^2 + 9x + 4

P(x) = (x - 2)(2x^2 + 9x + 4)

Q(x) can be factored solving

2x^2 + 9x + 4 = 0

x = - 1/2, x = -4

2x^2 + 9x + 4 = 2(x - (-1/2))(x - (-4)) = 2(x + 1/2)(x + 4) = (2x + 1)(x + 4)

putting all together

2x^3 + 5x^2 - 14x - 8 = (x - 2)(2x + 1)(x + 4)

if k = - 2, with a similar procedure you get

P(x) = - 2x^3 + 5x^2 + 14x - 8 = - (x + 2)(x - 4)(2x - 1)

You've asked it twice, but I'm afraid it's still impossible unless you know how to solve a quartic equation - Synthetic Division is no good since it doesn't factorise.

k | k_____5_____-7k_____-8

_ |______k²____k³ + 5k___k^4 - 2k²

_ | k___k² + 5___k³ - 2k___k^4 - 2k² - 8

k^4 - 2k² - 8 = 0

(k² - 4)(k² + 2) = 0

k = 2

2x³ + 5x² - 14x - 8