(2x^2+x-7)/(x-5) Long Division

3 min read Jun 16, 2024

Long Division of Polynomials: (2x^2 + x - 7) / (x - 5)

Long division is a fundamental concept in algebra used to divide polynomials. This article will guide you through the process of dividing the polynomial (2x^2 + x - 7) by (x - 5).

Setting up the Division

Arrange the polynomials: Write the dividend (2x^2 + x - 7) and the divisor (x - 5) in a long division format. Ensure both polynomials are written in descending order of their exponents.
```
     ________
x - 5 | 2x^2 + x - 7
```

Performing the Division

Divide the leading terms: Divide the leading term of the dividend (2x^2) by the leading term of the divisor (x). This gives us 2x.
```
     2x      
x - 5 | 2x^2 + x - 7 
```
Multiply the quotient by the divisor: Multiply the quotient (2x) by the divisor (x - 5) to get 2x^2 - 10x.
```
     2x      
x - 5 | 2x^2 + x - 7 
        2x^2 - 10x
```

Subtract: Subtract the result (2x^2 - 10x) from the dividend.

     2x      
x - 5 | 2x^2 + x - 7 
        2x^2 - 10x
        ---------
               11x - 7

Bring down the next term: Bring down the next term of the dividend (-7).

     2x      
x - 5 | 2x^2 + x - 7 
        2x^2 - 10x
        ---------
               11x - 7

Repeat steps 1-4: Now, divide the leading term of the new dividend (11x) by the leading term of the divisor (x), which gives us 11.
```
     2x + 11 
x - 5 | 2x^2 + x - 7 
        2x^2 - 10x
        ---------
               11x - 7
               11x - 55
```

Subtract: Subtract the result (11x - 55) from the previous result.

     2x + 11 
x - 5 | 2x^2 + x - 7 
        2x^2 - 10x
        ---------
               11x - 7
               11x - 55
               ---------
                      48

Result

The quotient of the division is 2x + 11 and the remainder is 48. We can express the result as:

(2x^2 + x - 7) / (x - 5) = 2x + 11 + 48/(x - 5)

Conclusion

This example demonstrates the process of polynomial long division. It is essential for simplifying polynomial expressions and solving algebraic equations. By understanding and practicing this technique, you can confidently manipulate and solve complex algebraic problems.