Questions & Answers

Question

Answers

$

\left( i \right){x^3} - 2{x^2} - x + 2 \\

\left( {ii} \right){x^3} + 13{x^2} + 32x + 20 \\

$

Answer
Verified

Hint: -To solve this type of factorization questions we have to add or subtract according to requirement that means we have to use our brain wisely to solve this type of questions as there is no definite method to proceed, different problems might require different adjustments.

Complete step by step answer:

We have

${x^3} - 2{x^2} - x - 2$

To make it in factor form we will do some rearrangement like,

$

{x^3} - {x^2} - {x^2} + x - 2x + 2 \\

{x^2}\left( {x - 1} \right) - x\left( {x - 1} \right) - 2\left( {x - 1} \right) \\

$

Now we will take ( x-1 ) common from all

$

\left( {x - 1} \right)\left( {{x^2} - x - 2} \right) \\

\left( {x - 1} \right)\left( {{x^2} - 2x + x - 2} \right) \\

\left( {x - 1} \right)\left( {x\left( {x - 2} \right) + 1\left( {x - 2} \right)} \right) \\

\left( {x - 1} \right)\left( {x - 2} \right)\left( {x + 1} \right) \\

$

Hence this is the factor form of the first question.

Now we have

$\left( {ii} \right){x^3} + 13{x^2} + 32x + 20$

Now we will make some rearrangement to make factor

$

{x^3} + {x^2} + 12{x^2} + 12x + 20x + 20 \\

{x^2}\left( {x + 1} \right) + 12x\left( {x + 1} \right) + 20\left( {x + 1} \right) \\

\\

$

Now taking ( x+1 ) common we get

$

\left( {x + 1} \right)\left( {{x^2} + 12x + 20} \right) \\

\left( {x + 1} \right)\left( {{x^2} + 10x + 2x + 20} \right) \\

\left( {x + 1} \right)\left( {x\left( {x + 10} \right) + 2\left( {x + 10} \right)} \right) \\

\left( {x + 1} \right)\left( {x + 10} \right)\left( {x + 2} \right) \\

$

Now this is the factor form of this question.

Note: -Whenever you get this type of question the key concept of solving is we have to do everything from our brain there is no any straight forward rule of solving factorization questions. Only you have to do exercise for easily solving factorization questions.

Complete step by step answer:

We have

${x^3} - 2{x^2} - x - 2$

To make it in factor form we will do some rearrangement like,

$

{x^3} - {x^2} - {x^2} + x - 2x + 2 \\

{x^2}\left( {x - 1} \right) - x\left( {x - 1} \right) - 2\left( {x - 1} \right) \\

$

Now we will take ( x-1 ) common from all

$

\left( {x - 1} \right)\left( {{x^2} - x - 2} \right) \\

\left( {x - 1} \right)\left( {{x^2} - 2x + x - 2} \right) \\

\left( {x - 1} \right)\left( {x\left( {x - 2} \right) + 1\left( {x - 2} \right)} \right) \\

\left( {x - 1} \right)\left( {x - 2} \right)\left( {x + 1} \right) \\

$

Hence this is the factor form of the first question.

Now we have

$\left( {ii} \right){x^3} + 13{x^2} + 32x + 20$

Now we will make some rearrangement to make factor

$

{x^3} + {x^2} + 12{x^2} + 12x + 20x + 20 \\

{x^2}\left( {x + 1} \right) + 12x\left( {x + 1} \right) + 20\left( {x + 1} \right) \\

\\

$

Now taking ( x+1 ) common we get

$

\left( {x + 1} \right)\left( {{x^2} + 12x + 20} \right) \\

\left( {x + 1} \right)\left( {{x^2} + 10x + 2x + 20} \right) \\

\left( {x + 1} \right)\left( {x\left( {x + 10} \right) + 2\left( {x + 10} \right)} \right) \\

\left( {x + 1} \right)\left( {x + 10} \right)\left( {x + 2} \right) \\

$

Now this is the factor form of this question.

Note: -Whenever you get this type of question the key concept of solving is we have to do everything from our brain there is no any straight forward rule of solving factorization questions. Only you have to do exercise for easily solving factorization questions.