DIVIDING POLYNOMIALS BY TRINOMIALS

Consider two polynomials f(x) and g(x), if we divide f(x) by g(x), we have to follow the given instruction.

Problem 1 :

(5x6 – 16x5 – 11x4 + 22x3 + 14x2 – 4x + 9) ÷ (x2 - 4x + 2)

Solution :

Let f(x) = 5x6 – 16x5 – 11x4 + 22x3 + 14x2 – 4x + 9

and

g(x) = x2 - 4x + 2

Step 1 :

Divide 5x6 by x2, so 5x6/x2 ==> 5x4

Put 5x4 at the top. Now, multiply 5x4 by (x2 - 4x + 2)

5x4 (x2 - 4x + 2) = 5x6 - 20x5 + 10x4

Write 5x6 - 20x5 + 10x4 below the given polynomial and then subtract.

Step 2 :

After subtraction, we will have a polynomial, take the first term from inside.

Here 4x5 should be divided by x2

4x5/x2 ==> 4x3

Now, multiply 4x3 by (x2 - 4x + 2)

 4x(x2 - 4x + 2) = 4x5 - 16x4 + 8x3

Subtract this polynomial from the previous one.

Repeat the process until, we receive degree of the remainder be lesser than to degree of dividend.

Note :

If the order of the given polynomial is rearranged, then write it in the correct order and start division.

IF any term is missing replace it by 0 and then divide.

Problem 1 :

(6x4 – 30x2 + 24) ÷ (2x2 – 8)

Solution :

(6x4 – 30x2 + 24) ÷ (2x2 – 8)

Quotient = 3x2 – 3

Remainder = 0

Problem 2 :

(3x5 + 4x3 – 5x + 8) ÷ (x2 + 3)

Solution :

(3x5 + 4x3 – 5x + 8) ÷ (x2 + 3)

Quotient = 3x3 – 5x

Remainder = 10x + 8

Problem 3 :

(x5 + 2x4 + x3 – x2 – 22x + 15) ÷ (x2 + 2x – 3)

Solution :

Quotient = x3 + 4x - 9

Remainder = 8x - 12

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