Question

For what value of ‘b’ is the polynomial x3 – 3x2 + bx – 6 divisible by x-3 ?


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Answer 1

f(x)=x3 - 3x2 + bx - 6

f(3)=27-27+3b-6

0=3b-6

3b=6

b=6/3

Therefore,

b=6

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Answer 2

Value of b is 2 .

If x³-3x²+bx-6 is divisible by x-3.

Given:

• Two polynomials.
• ${x}^{3} - 3 {x}^{2} + bx - 6$
• $x - 3$

To find:

• Value of b.If x³-3x²+bx-6 is divisible by x-3.

Solution:

Concept to be used:

If a polynomial is completely divisible by another polynomial then remainder should be zero.

Thus, Apply Remainder Theorem.

Remainder Theorem states that ,when p(x) is divided by (x-a), then remainder is p(a).

Step 1:

Apply Remainder Theorem.

Let

$p(x) = {x}^{3} - 3 {x}^{2} + bx - 6$

put x= 3

$p(3) = ( {3)}^{3} - 3( {3)}^{2} + b(3) - 6$

or

$p(3) = 27 - 27 + 3b - 6$

or

$\bf p(3) = 3b - 6$

Step 2:

As x-3 completely divided p(x), then remainder should equal to zero.

$p(3) = 0$

or

$3b - 6 = 0$

or

$3b = 6$

or

$b = \frac{6}{3}$

or

$\bf \red{b = 2}$

Thus,

Value of b is 2.

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