If x2 – 3x + 2 divides x3 – 6x2 + ax + b exactly, then find the value of ‘a’ and ‘b’
Answers
Answer:
⇒ Find the roots of :-
x² - 3x + 2
⇒ then place these values in the second polynomial ,you will get two equation ,solve them for calculating a and b.
x² - 3x + 2 = 0
x² - 2x - x + 2 = 0
x ( x-2 ) - 1 ( x-2 ) = 0
(x-2) (x-1) = 0
x - 2 = 0
x = 2
x - 1 = 0
x = 1
__________________
Put x = 2
x³ - 6x² + ax + b = 0
8 - 6(4) + 2a + b = 0
8 - 24 + 2a + b = 0
2a + b = 16
________________________
Put x = 1
x³ - 6x² + ax + b + 0
1 - 6 + a + b = 0
a + b = 5
_______________________
Subtract both the equations : -
2a - a = 16 - 5
a = 11
a + b = 5
11 + b = 5
b = 5 - 11
b = -6
kaisi hai ?
Answer:
The given polynomial is P(x)=2x
3
+3x
2
+ax+b
It is also given that if P(x) is divided by (x−2) then it will leave the remainder 2 and if divided by (x+2) then the remainder will be −2 which means that P(2)=2 and P(−2)=−2.
Let us first substitute P(2)=2 in P(x)=2x
3
+3x
2
+ax+b as shown below:
P(x)=2x
3
+3x
2
+ax+b
⇒P(2)=2(2)
3
+3(2)
2
+(a×2)+b
⇒2=(2×8)+(3×4)+2a+b
⇒2=16+12+2a+b
⇒2=28+2a+b
⇒2a+b=2−28
⇒2a+b=−26.........(1)
Now, substitute P(−2)=−2 in P(x)=2x
3
+3x
2
+ax+b as shown below:
P(x)=2x
3
+3x
2
+ax+b
⇒P(−2)=2(−2)
3
+3(−2)
2
+(a×−2)+b
⇒−2=(2×−8)+(3×4)−2a+b
⇒−2=−16+12−2a+b
⇒−2=−4−2a+b
⇒−2a+b=−2+4
⇒−2a+b=2.........(2)
Now adding the equations 1 and 2, we get
(2a−2a)+(b+b)=−26+2
⇒2b=−24
⇒b=−
2
24
⇒b=−12
Now substitute the value of b in equation 2:
−2a+(−12)=2
⇒−2a−12=2
⇒−2a=2+12
⇒−2a=14
⇒a=−
2
14
⇒a=−7
Hence a=−7 and b=−12