How do you add #2x ( 8x - 3) + ( 3x ^ { 2} + 6x - 4)#?

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Answer 1 · 2017-07-19T02:37:30

See a solution process below:

First, expand the term on the left by multiplying each term within the parenthesis by the term outside the parenthesis:

#color(red)(2x)(8x - 3) + (3x^2 + 6x - 4) =>#

#(color(red)(2x) * 8x) - (color(red)(2x) * 3) + (3x^2 + 6x - 4) =>#

#16x^2 - 6x + (3x^2 + 6x - 4)#

Next, remove all of the terms from parenthesis. Be careful to handle the signs of each individual term correctly:

#16x^2 - 6x + 3x^2 + 6x - 4#

Then, group like terms:

#16x^2 + 3x^2 - 6x + 6x - 4#

Now, combine like terms:

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

#19x^2 + 0x - 4#

#19x^2 - 4#

Answer 2 · 2017-07-19T02:39:23

See below

#19x^2-4#

To add #2x(8x−3)+(3x^2+6x−4)#

first distribute the #2x" "# #16x^2-6x +(3x^2+6x−4)#

next remove the parentheses (watch out for the sing if necessary

#16x^2-6x +3x^2+6x−4" "# and combine like terms

#19x^2-4#