Chapter ISG, Problem 1.3.7P

To determine

Solve the given equation $-2k+2K^2-3k+k^3$

Answer to Problem 1.3.7P

  $k\left(k^2+2k-5\right)$

Explanation of Solution

Given information:

  $-2k+2K^2-3k+k^3$

Formula used:

Simplification of equation

Calculation:

  $-2k+2K^2-3k+k^3$

Firstly arrange the all term

  $\left[\left(-2k+2k^2\right)-3k\right]+k^3$

Pull out like factor

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

Trying to factor by spliting the middle term

Factoring $\left(k^2+2k-5\right)$

The first term is $k^2$ its coefficient is 1

The middle term is +2k its coefficient is 2

The last term constant is 5

Multiply the coefficient of the first term by the constant

  $1\times \left(-5\right)=-5$

Find two factors of -5 whose sum equals the coefficient of the middle term,Which is 2

  $-5+1=-4$

Now second

  $-1+5=4$

No two such factors can be found

Hence, The final answer is

  $k\left(k^2+2k-5\right)$

Conclusion-

Trinomial can not be factored