Question

If two polynomiaus ax3+4x²+ 3x -4 & x3 - 4x + a leave the same remainder when divided by ( x - 3 ) , find the value of a​

Answer 1

❆ Given ❆

Two polynomials ax³ + 4x² + 3x - 4 and x³ - 4x + a leave the same remainder when divided by (x - 3).

❆ To Find ❆

The value of a​.

❆ Solution ❆

Zero of (x - 3) :-

x - 3 = 0

⇒x = 3

Let p(x) = ax³ + 4x² + 3x - 4 and f(x) = x³ - 4x + a.

A/q,

→ p(3) = f(3)

Putting the value of x = 3 :-

⇒a(3)³ + 4(3)² + 3(3) - 4 = (3)³ - 4(3) + a

⇒27a + (4 × 9) + 9 - 4 = 27 - 12 + a

⇒27a + 36 + 5 = 15 + a

⇒27a - a + 41 - 15 = 0

⇒26a + 26 = 0

⇒26a = -26

⇒a = -26/26

⇒a = -1

Answer 2

Solution -

Polynomial 1 => p ( x ) = ax³ + 4x² + 3x - 4

This is now divided by ( x - 3 ) .

Let us assume that the two polynomials when focused by ( x - 3 ) gives a remainder r in both cases .

So , p ( 3 ) of Polynomial 1 -

=> p ( 3 ) = a × 3³ + 4 × 3² + 3 × 3 - 4

=> p ( 3 ) = 27a + 36 + 9 - 4

=> p ( 3 ) = 27a + 41

So , r = 27a + 41

Now , Polynomial 2 is given as -

p ( y ) = y³ - 4y + a

This is now divided by ( x - 3 ) .

Let us assume that the two polynomials when focused by ( x - 3 ) gives a remainder r in both cases .

p ( 3 ) -

=> p ( 3 ) = 3³ - 4×3 + a

=> p ( 3 ) = 15 + a

So , r = 15 + a

But , r = 27a + 41

So , we can say that 15 + a and 27a + 41 are equal .

27a + 41 = a + 15

=> 26a = - 26

=> a = -1 .

Thus , the value of a comes out to be -1 .

This is the required answer .

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