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# Divide 15 (y + 3) (y² - 16) by 5 (y² - y - 12).

**Solution:**

__Factors__ of the term 15 (y + 3) (y² - 16) = 5 × 3 × ( y + 3) × (y - 4) × (y + 4) —(1)

Factors of the term 5 (y² - y - 12) = 5( y² - 4y + 3y - 12 ) = 5[y( y - 4)+ 3( y - 4) ] = 5(y - 4)(y + 3) —(2)

Now, dividing (1) by (2),

[15 (y + 3) (y² - 16)] / [5 (y² - y - 12)]

=[5 × 3 × ( y + 3) ×(y - 4)(y + 4) ] / [5(y - 4)(y + 3)]

= 3(y + 4)

**✦ Try This: **Divide 30(x+2)(x² -6x + 5) by 6(x² +x - 2)

Consider, x² -6x + 5 =x² - 5x - x + 5 = x( x - 5) - 1( x - 5) = (x -1)(x - 5)

Factors of the term: 30(x+2)(x² -6x + 5) = 2 x 3 x 5 x (x +2)(x - 5)( x - 1) ----(1)

Consider. x² + x -2 = x² +2 x - x -2 = x(x + 2)- 1( x + 2) = (x - 1)(x + 2)

Factors of the term: 6(x² + x -2) = 2 x 3 x (x -1 )(x + 2) -------(2)

Now, dividing (1) by (2)

[30(x+2)(x² -6x + 5)]/[ 6(x² +x - 2)]

= [2 x 3 x 5 x (x +2)(x - 5)( x - 1)] / [2 x 3 x (x -1 )(x + 2)]

= 5(x - 5)

**☛ Also Check: **NCERT Solutions for Class 8 Maths Chapter 9

**NCERT Exemplar Class 8 Maths Chapter 7 Sample Problem 17**

## Divide 15 (y + 3) (y² - 16) by 5 (y² - y - 12)

**Summary: **

Dividing 15 (y + 3) (y² - 16) by 5 (y² - y - 12) we get 3(y + 4)

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