Very difficult question....who can solve with right answers......?

Question

Show that the expression x^3 + (k-2)x^2 + (k-7)x - 4 has a factor x+1 for all values of k. If the expression also has a factor x+2.Find the value of k and the third factor.

Answer 1

1) Let p(x) = x^3 + (k - 2)x^2 + (k - 7)x - 4

Let's solve for p(-1).

p(-1) = (-1)^3 + (k - 2)(-1)^2 + (k - 7)(-1) - 4
p(-1) = -1 + (k - 2) - (k - 7) - 4
p(-1) = -1 + k - 2 - k + 7 - 4
p(-1) = 0

This implies that (-1) is a root, and it subsequently followed that
(x - (-1)), or (x + 1), is a factor.

2) If we're given (x + 2) is a factor, it follows that p(-2) must equal 0. That is,

p(-2) = (-2)^3 + (k - 2)(-2)^2 + (k - 7)(-2) - 4
p(-2) = -8 + 4(k - 2) - 2(k - 7) - 4
p(-2) = -8 + 4k - 8 - 2k + 14 - 4
p(-2) = 2k - 6

But, it is given that p(-2) = 0, so
2k - 6 = 0
2k = 6
k = 3

3) To find the third factor, we know it's going to be a real number. All we have to do is multiply the factors that we know
(x + 1)(x + 2), with the unknown factor, (x - r)

(x + 1)(x + 2)(x - r) = (x^2 + 3x + 2) (x - r)
= x^3 - rx^2 + 3x^2 - 3rx + 2x - 2r
= x^3 + (-r + 3)x^2 + (-3r + 2)x + (-2r)

All we have to do now is glance at the original equation, and match coefficients. Look at the coefficient of x^2, in this equation and in the original equation. We can EQUATE them.

k - 2 = -r + 3

But, we know the value of k; it's 3, so

3 - 2 = -r + 3

And we solve as normal.

1 = -r + 3
r = 2

Since r = 2 is a root it follows that (x - 2) is a factor.
So our third factor is (x - 2).

Answer 2

By the Factor Theorem, (x-n) is a factor of polynomial p(x) if n is a zero of p(x) - that is, if p(n)=0 - and vice versa. So plug -1 into your polynomial x^3 + (k-2)x^2 + (k-7)x - 4 and solve for k:

(-1)^3 + (k-2)(-1)^2 + (k-7)(-1) - 4 = 0
(-1) + k - 2 - k + 7 - 4 = 0

When you simplify this, you get 0 = 0, which is always true no matter what k is. Therefore, x+1 is always a factor.

To solve the second part, plug in x=-2:

(-2)^3 + (k-2)(-2)^2 + (k-7)(-2) - 4 = 0
(-8) + 4k - 8 - 2k + 14 - 4 = 0
2k - 6 = 0
k = 3

Therefore, your polynomial is x^3 + x^2 - 4x - 4. Factor out x+1:

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

Answer 3

for a no. to be a factor of a polynomial that expression equals=0.if f(x)=x^3+(k-2)x^2+(k-7)x-4 has a factor x+1 then
f(-1)=0.therefore
-1+k-2-k+7-4=0 since L.H.S =R.H.S THIS IS TRUE FOR EVERY VALUE OF K.(try putting any value of k for x=-1 you'll know it)
For x+2 to be factor f(-2)=0 therefore
-8+4k-8-2k+14-4=0
2k-6=0
k=3
for k=3 the polynomial will look like:x^3+x^2--4x-4
x^2(x+1)-4(x+1)=0
x+1(x^2-4)=0
hence the factors:-1,-2,2

Answer 4

x^3+(k-2)*x^2+(k-7)*x-4 = x^3+x^2-x^2+(k-2)*x^2 + (k-3-4)*x-4 = x^3+x^2 + (k-3)*x^2+(k-3)*x - (4*x+4) = (x+1)*(x^2+(k-3)*x-4)
so it has a factor of x+1 not depending of k
If it has a factor of x+2, then x^2+(k-3)*x-4 must have too a factor of x+2. Let's extract a multiple of x+2 and see what remains.
x^2+(k-3)*x-4 = x^2-2^2+(k-3)*x = (x+2)*(x-2)+(k-3)*x
So k=3 and the other factor is x-2

Answer 5

You said show the expression and the expression is x^3 + (k-2)x^2 + (k-7)x - 4

Answer 6

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