What is the standard form of a polynomial #(x-1)(x^2+6x+4)#?

1 Answer
Jun 12, 2017

Answer:

See a solution process below:

Explanation:

To put this expression in the standard form you must multiply these two terms by multiplying each individual term in the left parenthesis by each individual term in the right parenthesis.

#(color(red)(x) - color(red)(1))(color(blue)(x^2) + color(blue)(6x) + color(blue)(4))# becomes:

#(color(red)(x) xx color(blue)(x^2)) + (color(red)(x) xx color(blue)(6x)) + (color(red)(x) xx color(blue)(4)) - (color(red)(1) xx color(blue)(x^2)) - (color(red)(1) xx color(blue)(6x)) - (color(red)(1) xx color(blue)(4))#

#x^3 + 6x^2 + 4x - x^2 - 6x - 4#

We can now group and combine like terms:

#x^3 + 6x^2 - x^2 + 4x - 6x - 4#

#x^3 + 6x^2 - 1x^2 + 4x - 6x - 4#

#x^3 + (6 - 1)x^2 + (4 - 6)x - 4#

#x^3 + 5x^2 + (-2)x - 4#

#x^3 + 5x^2 -2x - 4#