How do you multiply $(3 x + 1) (x - 2)$ $(3x+1)(x-2)$ ?

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Answer 1 · 2015-07-04T03:38:09

Multiply each term in one factor time each term in the other factor

We need to multiply: $3 x$ $3x$ times $x$ $x$ and times $- 2$ $-2$ and we need to multiply $1$ $1$ times $x$ $x$ and times $- 2$ $-2$ .

We can (but are not required to) write:

$(3 x + 1) (x - 2) = (3 x \cdot x) + (3 x \cdot - 2) + (1 \cdot x) + (1 \cdot - 2)$ $(3x+1)(x-2) = (3x*x) + (3x * -2) + (1*x)+ (1*-2)$

We get:

$(3 x + 1) (x - 2) = 3 x^{2} + (- 6 x) + x + (- 2)$ $(3x+1)(x-2) = 3x^2 + (-6x) + x+ (-2)$

which is also equal to
$(3 x + 1) (x - 2) = 3 x^{2} - 6 x + x - 2$ $(3x+1)(x-2) = 3x^2 -6x + x -2$

Now simplify by combining like terms ( $- 6 x + x = - 5 x$ $-6x+x = -5x$ )

$(3 x + 1) (x - 2) = 3 x^{2} - 5 x - 2$ $(3x+1)(x-2) = 3x^2 -5x -2$

How do you multiply (3x+1)(x-2)(3x+1)(x−2)(3x+1)(x-2)?