Question

please help to solve this question of polynomial chapter ​

Answer 1

$\underline{\large{\tt{Question}}}$

x³ + 3x² + 3x + 1 is divided by ( x - a )

$\large\red{\frak{Solution}}$

Given the polynomial f (x) = x³ + 3x² + 3x + 1 and it is divided by ( x-a ) so as per remainder theorem

the remainder of the polynomial would be f ( a )

that is we got the remainder when we'll substitute the value of x and that is a since

• x - a = 0
• x = a

In this way the remainder would be

f(a) = a³ + 3a² + 3a + 1

Next step towards factorisation is that we have to find that coefficient value of a which when substituted in the above polynomial would bear a remainder 0 and that value must be observed as the factor of given polynomial f(x) as per the factor Theorem .

Thus , Let a = - 1

Substitute the value in f ( a )

$\tt f( - 1) = ( - 1)^{3} + 3( - 1)^{2} + 3( - 1) + 1 \\ = \cancel {- 1} + {\cancel { 3} - \cancel{ 3} + \cancel{ \: 1}} \red{ = 0}$

Therefore , ( a + 1 ) is a factor of the above polynomial.

To find the other factors of polynomial Let's divide the polynomial with its known factor.

$\begin{array} {c|c} \tt a + 1& \tt \: \cancel{a}^{3} + 3 {a}^{2} + 3a + 1\\&( - ) \tt \cancel {a}^{3} + {a}^{2} \\ \hline&( - )( - ) \\ &\tt \cancel{2 {a}^{2}} + 3a \\& \tt( - ) \cancel{2 {a}^{2}} + 2a \\ \hline& ( - )( - ) \\& \tt a + 1 \\ &( - ) \tt a + 1 \\ \hline&0 \\ \hline\end{array}$

On performing above division we get the remainder ( a² + 2a + 1 ) so this is the second factor of the polynomial.

Steps to show the factorisation :

a³ + 3a² + 3a + 1 = ( a + 1 )( a² + 2a + 1 )

We know

( a² + 2a + 1 ) = ( a + 1 )²

Therefore the factor of a³ + 3a² + 3a + 1 is ( a + 1 )³ .

$\rule{200pts}{3pts}$

$\large \frak{Factor}$

$= \tt(a + 1) ^{3}$

Thankyou