34 Chapter 2
Exercise 2.5
-2 2 0 -3 5 -7 -4 8 -10 10 Therefore Q 2 ( x ) = 2x -4 3 - 4x 5 2 + 5x -5 - 5 3
&
R = 3 Q2. Use synthetic division to find the value of k if -2 is a zero of
x 3 + 4x 2 + kx + 8 Sol: Since
x 3 + 4x 2 + kx +
8 Put in equation ( 2 ) we get Given zero x = -2 of polynomial
-2 1 4 K 8 -2 -4 -2k+8 1 2 k-4 -2k+16=R Since -2 is a zero of the given polynomial
Hence a =- 2, b = - 1 Or x + 2 is a factor so its Remainder = 0
Q5. One root of the cubic equation Therefore
R 2k 16 0
2k 16 k 16 2
8 is 3. Use synthetic division to find the other roots. Sol: Since Q3. Use synthetic division to find the value of
Or p and q if and are the factors of
3 1 0 -7 -6 3 9 6
Sol: Since
Therefore = - + = ⇒ - = - ⇒ = - - = x 1+ x 2- x 3 + px 2 + qx +
6 x 3 + px 2 + qx + 6 Factor x + 1 = 0 ⇒ x = -
1 1 3 2 0
and To find the other roots take Q (x) = 0 Factor x - 2 = 0 ⇒ x =
2 -1 1 P q 6 -1 -p+1 p-q-1 1 p-1 -p+q+1 p-q+5=0 2 2 2p+2 1 P+1 P+q+3=0 Since x + 1 and x -
2 are the factors of
Either or
the given polynomial p - q + 5 =
0 ( 1
)
so the remainder = 0
p + q + 3 =
0 ( 2
)
Hence the other roots are Q6. If – 1 and 2 are roots of the quartic 2p + 8 =
0
equation . Use 2p = -8synthetic division to find the other roots. p = - 2
8 = - 4 Put in eq ( 2 ) we get
Sol: Since
-1 1 -5 3 7 -2 -1 6 -9 2 - 4 + q + 3 =
0 q - 4 + 3 =
0
2 1 -6 9 -2 0
2 -8 2 1 -4 1 0 q - 1 =
0
Therefore q =
1
To find the other roots take Q (x) = 0 Hence
p = - 4,q =
1 Comparing with Q4. If x 1+ and x 2-
are the factors of the
we have 1 , - 4, 1 polynomial x 3 + ax 2 + bx +
2 , then using using quadratic formula synthetic division, find the values of a and b.
Putting values of a, b & c Sol: Since
x 3 + ax 2 + bx + 2 Factor x + 1 = 0 ⇒ x = - 1 Factor x - 2 = 0 ⇒ x =
2 -1 1 a b 2 -1 -a+1 a-b-1 1 a-1 -a+b+1 a-b+1=0 2 2 2a+2 1 a+1 a+b+3=0 Since x + 1 and x -
2 are the factors of
Hence the other roots are the given polynomial so the remainder = 0
2 - + b + 3 =
0 b - 2 + 3 =
0 b + 1 =
0 b = -
1
x 3- 7x - 6 =0 x 3- 7x - 6 =
0 x 3 + 0.x 2 - 7x - 6 =
0 Q ( x ) = x 2 + 3x + 2 R = 0 x 2+ 3x + 2 =
0 x 2+ 2x + 1x + 2 =
0 x ( x + 2 ) + 1 ( x + 2 )
=
0 ( x + 1 )( x + 2 )
=
0
x + 1 =
0 x = -
1
x + 2 =
0 x = -2
- 1, - 2 x 4 - 5x 3 + 3x 2 + 7x - 2 = 0 x 4 - 5x 3 + 3x 2 + 7x - 2 = 0 Q ( x ) = x 2 - 4x + 1 x 2- 4x + 1 = 0 ax 2 + bx + c = 0 a = b = c =
x =
- b ± b 2 - 4ac 2a
x =
- ( - 4 ) ± ( - 4 ) 2 - 4 ( 1 )( 1 )
2 ( 1 )x = 4 ± 2 16 - 4 =
4 ± 2 12 x = 4 ± 2 4 × 3 =
4 ± 2 2 3 x = 2 4 ± 2 2
3 = 2 ±3 2 + 3, 2 -
3
a - b + 1 =
0 1
a + b + 3 =
0 2
2a + 4 =
0 2a = -4a = - 2
4 = -
2
( ) ( )
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