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Find factors of x³-7x-6 A. (x-4)(x-2)(x+1) B. (x-6)(x-1)(x+1) C. (x-3)(x+2)(x+1) D. (x+3)(x+2)(x-1)

Respuesta :

Msurf617

Answer:

C. (x-3)(x+2)(x+1)

Step-by-step explanation:

We can use the rational roots test to help factor out the original equation.

The leading term is 1 and the constant is 6

p/q= 6/1

Now we find factors (all these are plus and minus)

1,2,3,6

1

We find the common ones (+1 and -1) and use -1 because it ends up being the root of the function

Factor, (x+1)

Now we have (x+1)(x^2-x-6)

Factor this with whatever method you perfer, I use AC method

Find two that are a product of -6 and add to -1 (-3 and 2)

We get (x+1)(x-3)(x+2)

C

ujalakhan18

Answer:

[tex]\boxed{C}[/tex]

Step-by-step explanation:

Let's solve all of the option and see which equals x³-7x-6

Option A)

[tex](x-4)(x-2)(x+1)[/tex]

=> [tex](x^2-6x+8)(x+1)[/tex]

=> [tex]x^3+x^2-6x^2-6x+8x+1\\x^3-5x^2+2x+1[/tex]

So, A is not correct

Option B)

[tex](x-6)(x-1)(x+1)\\(x+6)(x^2-1)\\x^3-x+6x^2-6\\x^2+6x^2-x-6[/tex]

This is also not correct

Option C)    ← Correct

[tex](x-3)(x+2)(x+1)\\(x^2-x-6)(x+1)\\x^3+x^2-x^2-x-6x-6\\x^3-7x-6[/tex]

This equals to x³-7x-6, So, this is the correct option. No need to do Option D since we have the right option now!

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