# Thread: X^3 + 2x^3 - 5x - 6 = (x-2)(x+3)(x-k), find k.

1. ## X^3 + 2x^3 - 5x - 6 = (x-2)(x+3)(x-k), find k.

X^3 + 2x^3 - 5x - 6 = (x-2)(x+3)(x-k), i have to find the value of k,

(ps of anyone is confuse ^ means power)

2. ## Re: Advance level past paper question

Expand the right-hand side and add like terms. Then use the fact that polynomials are equal iff their corresponding coefficients are equal.

3. ## Re: Advance level past paper question

Originally Posted by mathkid12
X^3 + 2x^3 - 5x - 6 = (x-2)(x+3)(x-k), i have to find the value of k,

(ps of anyone is confuse ^ means power)
Expand the RHS and equate the coefficients. A shortcut can be gained by considering the constant term: $\displaystyle -6 = -6 \cdot -k$

If you're unsure where -6 on the right comes from then what is the constant term of $\displaystyle (x-2)(x+3)$

4. ## Re: Advance level past paper question

I am assuming you mean:

$\displaystyle x^3+2x^2-5x-6=(x-2)(x+3)(x-k)$

In the factored form, look at the constant terms in each factor and what their product must be by looking at the expanded form.

5. ## Re: Advance level past paper question

Originally Posted by emakarov
Expand the right-hand side and add like terms. Then use the fact that polynomials are equal iff their corresponding coefficients are equal.

(x-2)(x+3)(x-k) = x^2 + 3x - 2x - 6 (x-k)
=x^3 + 3x^2 -6x^2 - 6x- kX^2 - 3xk + 2xk +6k

AM i goinG correct so far, and what advice can you give me to proceed, AND THANKS ALOT

6. ## Re: Advance level past paper question

Another way: $\displaystyle \frac{x^3+2x^2-5x+6}{(x-2)(x+3)}=\ldots= x+1$ so, $\displaystyle k=-1$ . You can quickly divide using Ruffini's rule.

7. ## Re: Advance level past paper question

ohh ok thanks but i still a bit confuse

8. ## Re: Advance level past paper question

If you are choosing to expand the RHS, then you will have:

$\displaystyle x^3+2x^2-5x-6=(x-2)(x+3)(x-k)$

$\displaystyle x^3+2x^2-5x-6=\left(x^2+x-6\right)(x-k)$

$\displaystyle x^3+2x^2-5x-6=\left(x^3+x^2-6x\right)-\left(kx^2+kx-6k\right)$

$\displaystyle x^3+2x^2-5x-6=x^3+(1-k)x^2-(6+k)x+6k$

Equating coefficients involving k, we find:

$\displaystyle 2=1-k$

$\displaystyle -5=-(6+k)$

$\displaystyle -6=6k$

Take your pick, and solve for k.

9. ## Re: Advance level past paper question

Originally Posted by mathkid12
X^3 + 2x^3 - 5x - 6 = (x-2)(x+3)(x-k), i have to find the value of k,

(ps of anyone is confuse ^ means power)

Find the value of k in terms of X right?

x^3 + 2x^3 - 5x -6 = (x-2)(x+3)(x-k)

= x^3 + 2x^3 - 5x - 6 = (x^2 + x - 6)(x - k)
= x^3 + 2x^3 - 5x - 6 = (x^3 - (k)x^2 + x^2 - (k)x -6x -(k)6
Subtract x^3 from both sides to get

2x^3 -5x - 6 = (k)x^2 + x^2 - (k)x - 6x - (k)6

Add 6x to both sides to get

2x^3 + x - 6 = (k)x^2 + x^2 - (k)x - (k)6

Subtract x^2 from both sides to get

2x^3 + x^2 + x - 6 = (k)x^2 + (k)x - (k)6

Take k out from the right side to get

2x^3 + x^2 + x - 6 = (k)(x^2 + x - 6) <--------- I may have messed up from this part, check your math too.

Look, we have x^2 + x - 6 on both sides so divide

2x^3 + 1 = (k)

10. ## Re: Advance level past paper question

Originally Posted by okokjae
Find the value of k in terms of X right?

x^3 + 2x^3 - 5x -6 = (x-2)(x+3)(x-k)

= x^3 + 2x^3 - 5x - 6 = (x^2 + x - 6)(x - k)
= x^3 + 2x^3 - 5x - 6 = (x^3 - (k)x^2 + x^2 - (k)x -6x -(k)6
Subtract x^3 from both sides to get

2x^3 -5x - 6 = (k)x^2 + x^2 - (k)x - 6x - (k)6

Add 6x to both sides to get

2x^3 + x - 6 = (k)x^2 + x^2 - (k)x - (k)6

Subtract x^2 from both sides to get

2x^3 + x^2 + x - 6 = (k)x^2 + (k)x - (k)6

Take k out from the right side to get

2x^3 + x^2 + x - 6 = (k)(x^2 + x - 6) <--------- I may have messed up from this part, check your math too.

Look, we have x^2 + x - 6 on both sides so divide

2x^3 + 1 = (k)
Scratch this post if you meant x^3 + 2x^2 - 5x - 6 = (x-2)(x+3)(x-k) and not x^3 + 2x^3 - 5x - 6 = (x-2)(x+3)(x-k)