Page 51 - SM inner class 8.cdr

3. Division of a polynomial by a binomial: (-4x ) by the first term of the
divisor (2x ) to get the second term
(i) Fac tor Method: If pos si ble, fac tor ise the
of the quotient (i.e. -4x ¸ 2x = -2)
given polynomial (which is to be divided) in
Step 5: Multiply the divisor (2x - ) 3 by -2
such a way that one of the fac tors is equal to
i.e. the second term of the quotient.
the binomial (by which we are dividing). We
Subtract the product -2 2( x - 3)
can cel the com mon fac tor and get the answer.
from (-4x + ) 6 i.e. -4x + 6 -
2
Example 17. Divide (x + 5 x + ) 6 by (x + 2 )
[-2 (2x - 3 )] = -4x + 6 + 4x - 6 = 0
2
Solution: First, factorise (x + 5 x + ) 6
3x - 2
2
x + 5 x + 6 = x 2 + x + x + 6
2
3
2
2x - 3 6x - 13x + 6
= x x + ) + 2 x ( + )
(
3
3
2
_6x m 9x
2
= (x + ) (x + )
3
_4x + 6
2
2
x + 5 x + 6 (x + ) (x + )
3
So, = m4x ± 6
2
x + 2 (x + )
0
Cancelling the common factor (x + 2 ) from the
numerator and denominator, we get. Quotient = 3x - 2
2
x + 5 x + 6 Remainder = 0
= x + 3
2
x + 2 The remainder is zero, Hence, (6x - 13x + ) 6
2
=
¸ (2x - ) 3x - 2. Hence, we see that 6x - 13x
3
(ii) Long Division Method: Remember that
(
(
the terms dividend, divisor, quotient and + 6 = 2x - 3) ´ 3x - 2) i.e.
remainder will also be used in the process of Dividend = Divisor ´ Quotient.
division of a polynomial by a polynomial as We find that (2x - ) 3 and (3x - ) 2 are the factors
you do in the case of division of a number by of (6x - 13x + ) 6
2
another number. Let us do an example to
explain the above method. The above example suggests the following
3
Example 18. Divide (6 13- x + 6x 2 ) by (- + 2x ) result:
If a polynomial p x( ) is divided by another
Step 1. Write the divisor (- +3 2x ) and
dividend (6 13- x + 6x 2 ) in polynomial r x( ) and leaves the remainder zero
(
decreasing order of the powers of and gives the quotient q x), then
(
)
(
(
the variable (i.e. in standard form) p x) = q x r x).
2
Dividend: 6x - 13x 6 Thus, q x( ) is a factor of p x( )
+
2
3
Divisor: 2x - 3 Example 19. Divide 8x + 10x - 6x + 3 by
1 + 4x and verify your answer.
Step 2: Divide the first term of the
dividend (6x 2 ) by the first term of Solution: We have standard form of
3
2
8x + 10x - 6x + 3 = 8x 3 - 6x 2 + 10x + 3 and
the divisor (2x ) to get the first
x
x
term of the quotient (3x ) (i.e. 1 + 4 = 4 + 1
2
6x ¸ 2x = 3x) 2x - 2x + 3
2
3
Step 3: Multiply the divisor (2x - 3) ) by 4x + 1 8x - 6x + 10x + 3
3
the first term of the quotient (3x ). _8x ± 2x 2
Subtract the product of (2x - ) 3 -8x 2 + 10x + 3
2
and (3x ) i.e. [3x (2x - ) 3 = 6x - 9x ] m 8x 2 m 2x
2
from the dividend (6x - 13x + ) 6 to
2
6
get the remainder [6x - 13x + - 12x + 3
2
(6x - 9x ) = - 4x + ] 6 _12x m 3
0
Step 4: Consider the remainder (-4x + ) 6
as the new dividend. Now divide Quotient = 2x 2 - 2x + 3, Remainder = 0
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Mathematics In Focus - 8