Is this correct
factor by grouping: x^3-5x^2+2x-10
My answer: (x^2+2)(x-5) *question do i need to go further?*
Errm dats almost done exept derz one more step, as you still need to chnage the quadratic equation into 2 linear ones as a Cubic equation has 3 linear functions, What i wud suggest is factorise da quadtracitic
it depends... as Mark said, you need to factor the first binomial(x^2 + 2), and if you require an answer: use the Zero Property of Multiplication to find out x sub 1 OR x sub 2.
Hope this helps...
Actually, the factorization you provided is not correct. Let's go through the correct process step by step.
To factor by grouping, you need to group the terms in pairs and look for common factors within each pair. Let's group the terms as follows:
(x^3 - 5x^2) + (2x - 10)
Now, we can factor out the greatest common factor from each pair separately.
From the first pair (x^3 - 5x^2), we can factor out x^2:
x^2(x - 5)
From the second pair (2x - 10), we can factor out 2:
2(x - 5)
Now, we can see that we have a common factor of (x - 5) between the two terms. We can factor it out:
(x - 5)(x^2 + 2)
Therefore, the correct factorization of the expression x^3 - 5x^2 + 2x - 10 is (x - 5)(x^2 + 2).
There is no need to go further, as this is the fully factored form.