
How do you factor \[{x^3} - 2{x^2} - 5x + 10?\]
Answer
555k+ views
Hint: irst use the commutative property and group first two and the second two terms, then find and take out the common terms between them and finally use the algebraic identity of difference between two square numbers to factorize further and get the desired factors. Algebraic identity for difference of two square numbers is given as
${a^2} - {b^2} = (a + b)(a - b)$
Formula used: Algebraic identity for difference between two square numbers: ${a^2} - {b^2} = (a + b)(a - b)$
Complete step-by-step solution:
In order to factorize the expression \[{x^3} - 2{x^2} - 5x + 10\] we will first group the terms with the help of commutative property,
\[ = \left( {{x^3} - 2{x^2}} \right) - \left( {5x - 10} \right)\]
Now taking out the common terms from the groups, we will get
\[ = {x^2}\left( {x - 2} \right) - 5\left( {x - 2} \right)\]
Taking $(x - 2)$ common in the above expression
\[ = \left( {x - 2} \right)\left( {{x^2} - 5} \right)\]
So do we get the required factors here? Yes we do get the factors but still the expression will further factorize one more time with the help of an algebraic identity.
We know the algebraic identity for difference between two square numbers that will be helpful to factorize the expression further as follows
In the expression
\[ = \left( {x - 2} \right)\left( {{x^2} - 5} \right)\]
\[\left( {{x^2} - 5} \right)\] can be written as \[\left( {{x^2} - {{\left( {\sqrt 5 } \right)}^2}} \right)\]
Now we can apply the algebraic identity of difference between two square numbers, here in the expression as follows
\[
= \left( {x - 2} \right)\left( {{x^2} - {{\left( {\sqrt 5 } \right)}^2}} \right) \\
= \left( {x - 2} \right)\left( {\left( {x + \sqrt 5 } \right)\left( {x - \sqrt 5 } \right)} \right) \\
= \left( {x - 2} \right)\left( {x + \sqrt 5 } \right)\left( {x - \sqrt 5 } \right) \\
\]
Therefore \[\left( {x - 2} \right),\;\left( {x + \sqrt 5 } \right)\;{\text{and}}\;\left( {x - \sqrt 5 } \right)\] are the required factors of the given expression.
Note: The number of factors of an expression depends upon the degree of that expression, if degree is two then its factors will also be two, if four then also four and if three as in this case then we can see that we get three factors of the given three degree expression.
${a^2} - {b^2} = (a + b)(a - b)$
Formula used: Algebraic identity for difference between two square numbers: ${a^2} - {b^2} = (a + b)(a - b)$
Complete step-by-step solution:
In order to factorize the expression \[{x^3} - 2{x^2} - 5x + 10\] we will first group the terms with the help of commutative property,
\[ = \left( {{x^3} - 2{x^2}} \right) - \left( {5x - 10} \right)\]
Now taking out the common terms from the groups, we will get
\[ = {x^2}\left( {x - 2} \right) - 5\left( {x - 2} \right)\]
Taking $(x - 2)$ common in the above expression
\[ = \left( {x - 2} \right)\left( {{x^2} - 5} \right)\]
So do we get the required factors here? Yes we do get the factors but still the expression will further factorize one more time with the help of an algebraic identity.
We know the algebraic identity for difference between two square numbers that will be helpful to factorize the expression further as follows
In the expression
\[ = \left( {x - 2} \right)\left( {{x^2} - 5} \right)\]
\[\left( {{x^2} - 5} \right)\] can be written as \[\left( {{x^2} - {{\left( {\sqrt 5 } \right)}^2}} \right)\]
Now we can apply the algebraic identity of difference between two square numbers, here in the expression as follows
\[
= \left( {x - 2} \right)\left( {{x^2} - {{\left( {\sqrt 5 } \right)}^2}} \right) \\
= \left( {x - 2} \right)\left( {\left( {x + \sqrt 5 } \right)\left( {x - \sqrt 5 } \right)} \right) \\
= \left( {x - 2} \right)\left( {x + \sqrt 5 } \right)\left( {x - \sqrt 5 } \right) \\
\]
Therefore \[\left( {x - 2} \right),\;\left( {x + \sqrt 5 } \right)\;{\text{and}}\;\left( {x - \sqrt 5 } \right)\] are the required factors of the given expression.
Note: The number of factors of an expression depends upon the degree of that expression, if degree is two then its factors will also be two, if four then also four and if three as in this case then we can see that we get three factors of the given three degree expression.
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