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Divide: 6x3 + 5x2 − 21x + 10 by 3x − 2 - Mathematics

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Question

Divide: 6x3 + 5x2 − 21x + 10 by 3x − 2 

Sum

Solution 1

6x3 + 5x2 − 21x + 10 by 3x − 2 

             2x2 + 3x − 5 
`"3x" − "2")overline("6x"^3 + "5x"^2 − 21"x" + 10)(`
             6x3 − 4x2 
            −      +           
          9x2  − 21x 
          9x2  − 6x 
        −          +            
              − 15x + 10
              − 15x + 10
             +        −         
           xxx              
= 2x2 + 3x − 5 

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Solution 2

`(6x^3 + 5x^2 - 21x + 10)/(3x-2)`

Divide the leading term of the numerator (6x3) by the leading term of the denominator (3x): `(6x^2)/(3x) = 2x^2`

(3x − 2) × 2x2 = 6x3 − 4x2

Subtract 6x3 − 4x2 from the original numerator:

(6x3 + 5x2 − 21x + 10) − (6x3 − 4x2) = 9x2 − 21x + 10.

Divide the leading term of the new expression (9x2) by the leading term of the divisor (3x): `(9x^2)/(3x) = 3x`

(3x − 2) × 3x = 9x2 − 6x.

Subtract 9x2 − 6x from the current expression:

(9x2 − 21x + 10) − (9x2 − 6x) = −15x + 10

Divide the leading term of the new expression (−15x) by the leading term of the divisor (3x): `(-15x)/(3x) = -5`

(3x − 2) × − 5 = −15x + 10

Subtract −15x + 10 from the current expression:

(−15x + 10) − (−15x + 10) = 0.

The quotient is: 2x2 + 3x - 5

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Chapter 11: Fundamental Concepts (Including Fundamental Operations) - Exercise 11 (D)

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Selina Concise Mathematics [English] Class 7 ICSE
Chapter 11 Fundamental Concepts (Including Fundamental Operations)
Exercise 11 (D) | Q 2.12
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