Question

if x^3+ax^2-bx+10 is divisible by x^2-3x+2 then find the value of a and b

Answer 1

Hey.

Here is the answer.

As x^3+ax^2-bx+10 is divisible by x^2-3x+2 then x^2-3x+2 =(x-1)(x-2) is factor.

So, x=1 and 2 are zeros of the given polynomial .

Putting x=1

1^3+a×1^2-b×1+10=0

or, 1+a-b+10=0

or, a-b= -11___________1st


Putting x=2

2^3+a×2^2-b×2+10=0

or, 8+4a-2b+10=0

or, 2a-b= -9____________2nd

Solving 1st & 2nd , we get

a= 2

b= 13

Thanks.

Answer 2

x² - 3x + 2
=> x² - (2 + 1)x + 2
=> x² - 2x - x + 2
=> x(x - 2) - 1(x - 2)
=> (x - 2)(x - 1)

========================
Then, we get x= 1 or 2

Putting the value of x given equation, [taking 1]

x³ + ax² - bx + 10 = 0

1³ + a(1)² - b(1) + 10 = 0

1 + a - b + 10 = 0

a - b = -11 ----1equation

××××××××××××××××

Taking x = 2,

2³ + a(2)² - b(2) + 10=0

8 + 4a - 2b + 10 = 0

18 -2(-2a + b) = 0

-2(-2a + b) = -18

-2a + b = 9

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

Adding both the equations,

a - b = -11
-2a + b = 9
_________
-a = -2
__________

a = 2

Putting the value of a in 1equation,

a - b = -11

2 - b = -11

2 + 11 = b

13 = b



I hope this will help you

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