Division of Algebraic Expressions Continued ( Polynomial ÷ Polynomial)

When we a polynomial in both the numerator and the denominator, we factorise them both and bring them down to their irreducible forms and accordingly check for any common factors(like earlier).

Solve 7x2+14x÷x+2

Factorise7x2+14x

Factorised form of numerator:
Now, 7x2+14x÷x+2 =
Thus, the quotient is 7x after cancelling the common factor: (x+2)

Let's Solve

Factorise and solve the following:

5p2−25p+20÷p−1

Factorised form of numerator:
Now let's check our answer, taking out the common factor of 5, we get: 5p2−5p+4
This can be re-written as: 5p2−5p+4=5p2−p−4p+4=5p−1p−4
Now, 5p2−25p+20÷p−1 =
Thus, the quotient is 5×p−4 after cancelling the common factor: (p-1)

39y350y2−98÷26y25y+7

Factorise 39y350y2−98

Factorised form of numerator:
Now let's check our answer, taking out the common factors of 2, 39 and y3, we get: 5y−75y+7
This gives us: 39y350y2−98=78×y3×5y+75y−7
Now, 39y350y2−98÷26y25y+7 =
Thus, the quotient is 3y5y−7 after cancelling the common factors: 26, y2 and 5y+7

12xy9x2−16y2÷4xy3x+4y

Factorise 12xy9x2−16y2

Factorised form of numerator:
Using the identity- a2−b2=a+ba−b, we get: 3x+4y3x−4y
This gives us: 12xy9x2−16y2=12xy3x+4y3x−4y
Now, 12xy9x2−16y2÷4xy3x+4y =
Thus, the quotient is 33x−4y after cancelling the common factors: x, y and 3x+4y

4yzz2+6z−16÷2yz+8

Factorise 4yzz2+6z−16

Factorised form of numerator:
On factorising z2+6z−16, we get: z+8z−2
This gives us: 4yzz2+6z−16=4yzz−2z+8
Now, 4yzz2+6z−16÷2yz+8 =
Thus, the quotient is 2zz−2 after cancelling the common factors: y and z+8

Solve the following division problems:

(a) −36y3÷9y2