33 Chapter 2
Exercise 2.5
Remainder Theorem: A polynomial of degree n ≥ 1 , divided by ( x - a ) until a constant is
obtained called remainder denoted by f ( a ) Factor Theorem: A polynomial f ( x ) has a factor ( x - a ) iff f ( a ) = 0 Synthetic Division: The method of synthetic division for dividing the polynomial of degree n ax n n + a n - 1 x n - 1 + a n - 2 x n
-
2 + + ax 1
1
+
a o by x a- is illustrated and described as follow: a a n a n - 1 a n -
2 a 1
a o
Row one
↓
aa + n
ab + ac + ad + ae + Row
Two
a n b c d e f Row
Three Re mainder Coefficients of Quotient
where
and e = a 1+
ad f = a o
+
ae
Quotient should be a nx n - 1 + bx n - 2 + cx n
- 3 + + dx +
e of degree ( n- 1) if Remainder is zero or f = 0 And Quotient is also known as Depressed polynomial. Exp19: Use synthetic division to find the Quotient Q(x) and the remainder R when 3 x 3 - 2 x 2 - 150 is divided by x- 4 Sol: Given P ( x ) = 3 x 3 - 2 x 2 + 0 x - 150 And divisor x- 4 = 0 ⇒ x = 4 Then by synthetic division
4 3 -2 0 -150 12 40 160 3 10 40 10 Therefore Q(x) = 3 x 2 + 10 x + 40 & R = 10 Exp20: use synthetic division to find the value of k if 2 is a zero of 2 x 4 + x 3 + kx 2 - 8 Sol: Given P ( x ) = 2 x 4 + x 3 + kx 2 + 0 x - 8 And given zero is 2 Then by synthetic division
2 2 1 K 0 -8 4 10 2k+20 4k+40 2 5 k+10 2k+20 4k+32 Since 2 is a zero of Polynomial so
4 k+ 32 =
0 4 k= -
32 k
= -
8 Exp21: use synthetic division to find the value of m and n if x- 1 & x+ 2 are factors of
x 3 - mx 2 + nx + 12 Sol: Given P ( x ) = x 3 - mx 2 + nx + 12 Given x- 1 and x+ 2 are factors. There zeros x - 1 = 0 ⇒ x = 1 & x + 2 = 0 ⇒ x = - 2 Then by synthetic division
1 1 -m n 12 1 1-m 1-m+n 1 1-m 1-m+n 13-m+n -2 -2 2+2m 1 -1-m 3+m+n
b = a n -1+
aa
n c = a n
-
2
+
ab d = +
ac
Since 1 &-2 is a zero of Polynomial so we have
13 - m + n = 0 .............(1) 3 + m + n = 0 ............(2) 16 + 2 n = 0 by adding eq(1) and eq (2) n⇒ = - 8 put in eq (2) 3 + m
+ ( - 8 ) =
0 m- 5 =
0 m= 5Thus m= 5 & n = - 8 Exp22:if -1 and 2 are roots of quartic equation x 4 - 5 x 2 + 4 = 0 use synthetic division to find other roots Solution: Given quartic eq x 4 - 5 x 2 + 4 = 0 And roots are -1 and 2 Then by synthetic division -1 1 0 -5 0 4 -1 1 4 -4 1 -1 -4 4 0 2 2 2 -4 1 1 -2 0 Q ( x ) = x 2 + x - 2 = 0 having other factors
x 2 + 2 x - 1 x
- 2 =
0 x ( x + 2 ) - 1 ( x
+ 2 )
= 0 ( x + 2 )( x
- 1 )
=
0 Either x+ 2 = 0 or x- 1 =
0 x = - 2 x = 1 Thus other two roots are x = - 2 & x =
1 Exercise 2.5 Q1i). Use synthetic division to find the quotient and remainder R when
is divided by Sol: Since And -3 3 2 -1 -1 -9 21 -60 3 -7 20 -61 Therefore &
Q1ii). Use synthetic division to find the quotient and remainder R when
is divided by Sol: Since And 3 2 -7 12 -27 6 -3 27 2 -1 9 0 Therefore and
Q1iii). Use synthetic division to find the quotient and remainder R when
is divided by Sol: Since
And
Q ( x ) 3x 3 + 2x 2 - x - 1 x 3+ 3x 3 + 2x 2 - x - 1 x + 3 = 0 ⇒ x = -
3 Q ( x ) = 3x 2 - 7x + 20 R = - 61 Q ( x ) 2x 3 - 7x 2 + 12x - 27 x 3- 2x 3 - 7x 2 + 12x - 27 x - 3 = 0 ⇒ x =
3 Q ( x ) = 2x 2 - x + 9 R = 0 Q ( x ) 2x 4 - 3x 2 + 5x - 7 x 2+ 2x 4 - 3x 2 + 5x - 7 = 2x 4 + 0.x 3 - 3x 2 + 5x -
7 x + 2 = 0 ⇒ x = -
2
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