Question

Find the value of a and b so that p(x)=x3-10x2+ax+b is divisible by x-1 as well as x-2

Answer 1

Heya !!!

(X -1) and (X-2) are the two zeroes of the given polynomial.

X -1 = 0

X = 1

And,

X -2 = 0

X = 2

P(X) = X³ - 10X² + AX + B

P(1) = (1)³ - 10 × (1)² + A × 1 + B

=> 1 - 10 × 1 + A + B = 0

=> 1 - 10 + A + B = 0

=> A + B = 9 --------(1)

Also X = 2

P(2) = (2)³ - 10 × (2)² + A × 2 + B

=> 8 - 10 × 4 + 2A + B

=> 8 - 40 + 2A + B = 0

=> 2A + B = 32 ------(2)

From equation (1) we get,

A + B = 9


A = 9 - B


Putting the value of B in equation (2) we get,


2A + B = 32


2 × (9-B) + B = 32


18 - 2B + B = 32



-B = 32 -18


-B = 14


B = -14


Putting B = -14 in equation (3).


A = 9 - B => 9 - (-14)


A = 9 +14 = 23

Hence,

A = 23 and B = -14.

HOPE IT WILL HELP YOU...... :-)

Answer 2

One factor, x - 1

x - 1 = 0
x = 1

---------

Taking x as 1,

x³ - 10x² + ax + b = 0

=> 1³ - 10(1)² + a(1) + b(1) = 0

=> 1 - 10 + a + b =0

=> -9 +a + b = 0

=> a + b = 9 -----1equation

×××××××××××××××××

Other factor, x - 2

x - 2 = 0
x = 2

----------
Taking x as 2

x³ - 10x² + ax + b = 0

=> (2)³ - 10(2)² + a(2) + b = 0

=> 8 - 40 + 2a + b = 0

=> -32 + 2a + b = 0

=> 2a + b = 32 -------2equation

××××××××××××××××××××××
Subtract eq(2) from eq(1),

a + b = 9
2a + b = 32
(-)_(-) __(-)
-a = -23
_________

a = 23

==================

Putting the value of a in 1equation,

a + b = 9
23 + b = 9
b = 9 - 23
b = -14

=========================

Then, a = 23 and b = -14

I hope this will help you

. (-: