
How to factor \[{x^3} - 8{x^2} + 15x\]?
Answer
446.4k+ views
Hint:
Here, we will factorize the cubic equation such that it will become a combination of a linear equation and a quadratic equation. We will find the factors of the quadratic equation by using the middle terms split method. Then by combining the factors of both the equations, we will get the factors of the given cubic equation.
Complete step by step solution:
We are given a cubic equation \[{x^3} - 8{x^2} + 15x\].
We will now find the factors of the given cubic equation.
First, we will take out the factor common to all the terms, so we get
\[ \Rightarrow {x^3} - 8{x^2} + 15x = x\left( {{x^2} - 8x + 15} \right)\]……………………………………………………………………..\[\left( 1 \right)\]
Now, we have a quadratic equation \[{x^2} - 8x + 15\]. So, we will find the factors of the quadratic equation.
We know that the general form of a Quadratic equation is \[a{x^2} + bx + c\].
Comparing the given quadratic equation with the general quadratic equation, we get
\[\begin{array}{l}a = 1\\b = - 8\\c = 15\end{array}\]
Now, in order to factorize the quadratic equation, we will find a pair of numbers, such that the product of the numbers is equal to the product of the first term and third term. Also, the sum of the numbers is equal to the middle terms of the equation.
Thus, the numbers \[ - 3\] and \[ - 5\] satisfies the above mentioned condition as their Product is equal to \[ac = \left( 1 \right)\left( {15} \right) = 15\] and their sum is equal to the middle terms, \[ - 8\].
Now, we will split the middle term in the given quadratic equation as \[ - 3x\] and\[ - 5x\] . So, we get
\[{x^2} - 8x + 15 = {x^2} - 3x - 5x + 15\].
By taking out the common factors, we get
\[ \Rightarrow {x^2} - 8x + 15 = x\left( {x - 3} \right) - 5\left( {x - 3} \right)\].
By grouping the common factors, we get
\[ \Rightarrow {x^2} - 8x + 12 = \left( {x - 3} \right)\left( {x - 5} \right)\].
Now, by substituting the factors of the quadratic equation in equation \[\left( 1 \right)\] , we get
\[{x^3} - 8{x^2} + 15x = x\left( {x - 3} \right)\left( {x - 5} \right)\] .
Therefore, the factors of \[{x^3} - 8{x^2} + 15x\] are\[x\] ,\[\left( {x - 3} \right)\] and \[\left( {x - 5} \right)\].
Note:
We know that the cubic equation is defined as an equation with the highest degree as 3. We know that Factorization is a process of rewriting the expression in terms of the product of the factors. Factorization is done by using the common factors, the grouping of terms and the algebraic identity. We should remember that if the sign of the product of the factors is Positive, then both the integers should be either positive or negative and the sign of the sum of the factors is Negative, then both the integers should be negative.
Here, we will factorize the cubic equation such that it will become a combination of a linear equation and a quadratic equation. We will find the factors of the quadratic equation by using the middle terms split method. Then by combining the factors of both the equations, we will get the factors of the given cubic equation.
Complete step by step solution:
We are given a cubic equation \[{x^3} - 8{x^2} + 15x\].
We will now find the factors of the given cubic equation.
First, we will take out the factor common to all the terms, so we get
\[ \Rightarrow {x^3} - 8{x^2} + 15x = x\left( {{x^2} - 8x + 15} \right)\]……………………………………………………………………..\[\left( 1 \right)\]
Now, we have a quadratic equation \[{x^2} - 8x + 15\]. So, we will find the factors of the quadratic equation.
We know that the general form of a Quadratic equation is \[a{x^2} + bx + c\].
Comparing the given quadratic equation with the general quadratic equation, we get
\[\begin{array}{l}a = 1\\b = - 8\\c = 15\end{array}\]
Now, in order to factorize the quadratic equation, we will find a pair of numbers, such that the product of the numbers is equal to the product of the first term and third term. Also, the sum of the numbers is equal to the middle terms of the equation.
Thus, the numbers \[ - 3\] and \[ - 5\] satisfies the above mentioned condition as their Product is equal to \[ac = \left( 1 \right)\left( {15} \right) = 15\] and their sum is equal to the middle terms, \[ - 8\].
Now, we will split the middle term in the given quadratic equation as \[ - 3x\] and\[ - 5x\] . So, we get
\[{x^2} - 8x + 15 = {x^2} - 3x - 5x + 15\].
By taking out the common factors, we get
\[ \Rightarrow {x^2} - 8x + 15 = x\left( {x - 3} \right) - 5\left( {x - 3} \right)\].
By grouping the common factors, we get
\[ \Rightarrow {x^2} - 8x + 12 = \left( {x - 3} \right)\left( {x - 5} \right)\].
Now, by substituting the factors of the quadratic equation in equation \[\left( 1 \right)\] , we get
\[{x^3} - 8{x^2} + 15x = x\left( {x - 3} \right)\left( {x - 5} \right)\] .
Therefore, the factors of \[{x^3} - 8{x^2} + 15x\] are\[x\] ,\[\left( {x - 3} \right)\] and \[\left( {x - 5} \right)\].
Note:
We know that the cubic equation is defined as an equation with the highest degree as 3. We know that Factorization is a process of rewriting the expression in terms of the product of the factors. Factorization is done by using the common factors, the grouping of terms and the algebraic identity. We should remember that if the sign of the product of the factors is Positive, then both the integers should be either positive or negative and the sign of the sum of the factors is Negative, then both the integers should be negative.
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