How do you Factor completely \[2{x^3} + 5{x^2} - 37x - 60\]?
Answer
622.5k+ views
Hint: Here, we will find the factors by separating the middle terms of the cubic equation. Then we will factor out common terms and get the factors in terms of linear expression and quadratic expression. We will then factorize the quadratic expression using the middle term splitting method to get the required answer.
Complete step by step solution:
We are given a function \[f\left( x \right) = 2{x^3} + 5{x^2} - 37x - 60\].
Now, we will split the middle terms of the expansion, we get
\[ \Rightarrow f\left( x \right) = 2{x^3} - 8{x^2} + 13{x^2} - 52x + 15x - 60\].
Now, we will take out the common factors in the expansion, we get
\[ \Rightarrow f\left( x \right) = 2{x^2}\left( {x - 4} \right) + 13x\left( {x - 4} \right) + 15\left( {x - 4} \right)\].
Now, we will take out the common factors, we get
\[ \Rightarrow f\left( x \right) = \left( {x - 4} \right)\left( {2{x^2} + 13x + 15} \right)\].
Now, we will find the factors by using the factorization, we get
\[ \Rightarrow f\left( x \right) = \left( {x - 4} \right)\left( {2{x^2} + 10x + 3x + 15} \right)\].
Now, we will take out the common factors in the expansion, we get
\[ \Rightarrow f\left( x \right) = \left( {x - 4} \right)\left( {2x\left( {x + 5} \right) + 3\left( {x + 5} \right)} \right)\].
Now, we will take out the common factors, we get
\[ \Rightarrow f\left( x \right) = \left( {x - 4} \right)\left( {2x + 3} \right)\left( {x + 5} \right)\].
Therefore, the factors of \[2{x^3} + 5{x^2} - 37x - 60\] is \[\left( {x - 4} \right)\left( {2x + 3} \right)\left( {x + 5} \right)\].
Note:
We know that Factorization is a process of rewriting the expression in terms of the product of the factors. Factors are numbers if the expression is a numeral. Factors are algebraic expressions if the expression is an algebraic expression. 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 positive, then both the integers should be positive. This determines the sign of the factors.
Complete step by step solution:
We are given a function \[f\left( x \right) = 2{x^3} + 5{x^2} - 37x - 60\].
Now, we will split the middle terms of the expansion, we get
\[ \Rightarrow f\left( x \right) = 2{x^3} - 8{x^2} + 13{x^2} - 52x + 15x - 60\].
Now, we will take out the common factors in the expansion, we get
\[ \Rightarrow f\left( x \right) = 2{x^2}\left( {x - 4} \right) + 13x\left( {x - 4} \right) + 15\left( {x - 4} \right)\].
Now, we will take out the common factors, we get
\[ \Rightarrow f\left( x \right) = \left( {x - 4} \right)\left( {2{x^2} + 13x + 15} \right)\].
Now, we will find the factors by using the factorization, we get
\[ \Rightarrow f\left( x \right) = \left( {x - 4} \right)\left( {2{x^2} + 10x + 3x + 15} \right)\].
Now, we will take out the common factors in the expansion, we get
\[ \Rightarrow f\left( x \right) = \left( {x - 4} \right)\left( {2x\left( {x + 5} \right) + 3\left( {x + 5} \right)} \right)\].
Now, we will take out the common factors, we get
\[ \Rightarrow f\left( x \right) = \left( {x - 4} \right)\left( {2x + 3} \right)\left( {x + 5} \right)\].
Therefore, the factors of \[2{x^3} + 5{x^2} - 37x - 60\] is \[\left( {x - 4} \right)\left( {2x + 3} \right)\left( {x + 5} \right)\].
Note:
We know that Factorization is a process of rewriting the expression in terms of the product of the factors. Factors are numbers if the expression is a numeral. Factors are algebraic expressions if the expression is an algebraic expression. 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 positive, then both the integers should be positive. This determines the sign of the factors.
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