Question

x-2 × 2x²+x-3 + -8
pls answer it it's urgent
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Answer 1

Given,

$(x-2)(2x^{2} +x-3)+(-8)$

• First we have to multiply $(x-2)(2x^{2} +x-3)$ by using distribution property of multiplication.
• Distribution property of multiplication: It states that multiplying two factors together gives the same result as breaking one factor up into two addends, multiplying both addends with the remaining factor, then adding both products together.

$= x(2x^{2} +x-3)-2(2x^{2} +x-3)+(-8)$

$=2x^{3}+x^{2} -3x-(4x^{2} +2x-6)-8$

$=2x^{3}+x^{2} -3x-4x^{2} -2x+6-8$

$=2x^{3}-3x^{2} -5x-2$

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

Therefore, the answer is $x^{2}(2x-3)-5x-2$.

Answer 2

Answer:

To solve : $( x - 2 ) ( 2x^{2} + x - 3 ) + (- 8)$

To solve given expression -

Using  distribution law of multiplication

when we multiply a number with the sum of two or more addends or minuends, result obtained  is equal to the result that is obtained when we multiply each addend or minuend separately by the number.

hence , on solving $( x - 2 ) ( 2x^{2} + x - 3 ) + (- 8)$

we get,  $= 2x^{3} + x^{2} - 3x -4x^{2} -2x + 6 - 8$

             $= 2x^{3} - 3x^{2} - 5x - 2$

hence, $2x^{3} - 3x^{2} - 5x - 2$ is the required answer.

           

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