Question

How do i distribute (x+3](2x-1)(-2x+1)?

Answer 1

There are two ways to execute this equational expression. One is without the use of distributive law, which is going to be a long process given, we are not skipping the middle steps and solving it fully. Second one is about using distributive law. Both the methods will yield the same answer as a fully distributed expression. So, here is my process via LaTeX, the equation editor to interpret mathematical expressions.

Method 1: Without the use of distributive law.

Just apply the FOIL method which is given by the rule of expanding the brackets.


`\boxed{\mathbf{FOIL \: Rule: \: (a + b) (c + d) = ac + ad - bc + bd}}`


`\mathbf{x * 2x + x (- 1) + 3 * 2x + 3 (- 1) \big(- 2x + 1 \big)}`


`\mathbf{2xx - 1 * x + 3 * 2x - 3 * 1 \big(- 2x + 1 \big)}`


`\mathbf{\big(2x^2 - x + 6x - 3 \big) \big(- 2x + 1 \big)}`


`\mathbf{\big(2x^2 - 5x - 3 \big) \big(- 2x + 1 \big)}`


`\mathbf{2x^2 (- 2x) + 2x^2 * 1 + 5x (- 2x) + 5x * 1 + (- 3) (- 2) + (- 3) * 1}`


`\mathbf{- 2 * 2x^2x + 2 * 1 * x^2 - 5 * 2xx + 5 * 1 * x + 3 * 2x - 3 * 1}`


`\mathbf{- 4x^3 + 2x^2 - 10x^2 + 5x + 6x - 3}`


`\mathbf{- 4x^3 + 2x^2 - 10x^2 + 11x - 3}`


`\boxed{\mathbf{\underline{Final \: \: Answer: - 4x^3 - 8x^2 + 11x - 3}}}`

The second method relies on variable values equalling to the numbered values, it is a little more longer process than the former one. I advise you to stick to this method for now as a distribution for the bracketed expressions.

Hope it helps.