Question

If x-a is a factor of $p\left( x \right)={{x}^{3}}-3{{x}^{2}}a+2{{a}^{2}}x+b$, the the value of b is
[a] 0
[b] 2
[c] 1
[d] 3

Answer 1

Hint: Use factor theorem. According to factor theorem, if x-a is a factor of p(x), then p(a) = 0. Use factor theorem and hence form an equation in b. Solve for b. The value of b gives the value such that x-a is a factor of p(x).

Complete step-by-step answer:
Alternatively, use synthetic division to determine at what value of b is x-a a factor of p(x).

We have $p\left( x \right)={{x}^{3}}-3{{x}^{2}}a+2{{a}^{2}}x+b$.
Since x-a is a factor of p(x), we have
$p\left( x \right)=\left( x-a \right)g\left( x \right)$ where g(x) is another polynomial of lower degree.
Put x = a, we get
$p\left( a \right)=\left( a-a \right)g\left( a \right)=0$
Hence, we have
p(a) =0.
Now, we have
$p\left( x \right)={{x}^{3}}-3{{x}^{2}}a+2{{a}^{2}}x+b$
Put x = a in the expression of p(x), we get
$p\left( a \right)={{\left( a \right)}^{3}}-3{{a}^{2}}a+2{{a}^{2}}a+b={{a}^{3}}-3{{a}^{3}}+2{{a}^{3}}+b=b$
Hence, p(a) = b
But p(a) = 0
Hence, we have
b =0
Hence the value of b at which x-a is a factor of p(x) is 0.
Hence option [a] is correct.

Note: Alternative solution: Best Method: Synthetic division.
In this method, we start by writing coefficients of the polynomial in order from the highest degree to the constant term. If in between some degree terms are missing, we set their coefficient as 0.
Hence $p\left( x \right)={{x}^{3}}-3a{{x}^{3}}+2{{a}^{2}}x+b$ will be written as

 

Now each the point which has to be substituted(say x= a) is written as follows

0 is placed below the first term

Now the terms under the same column are added. The sum is then multiplied with the root, and the product is written under the coefficient of the next term.
Hence, we have

Continuing in this way we have the following

Since the last sum is b, we have $p\left( a \right)=b$.
Hence, b = 0 by factor theorem.