
How do you factor the trinomial \[6{{x}^{2}}+13x+5\]?
Answer
563.4k+ views
Hint:
A trinomial is a polynomial with only three terms. If \[a{{x}^{2}}+bx+c\] is a trinomial, when it does not factor into a square, we factorize it by extension of factoring. That is, we use the sum-product method to factor the trinomial. We can check the factor trinomial by the fact that when simplified it is the same as the actual trinomial.
Complete step by step answer:
For a trinomial expression like \[a{{x}^{2}}+bx+c\], we try to write the c term as the product of two numbers, and b term as the sum (or difference) of the same two numbers. In the case of both being positive, it’s not hard to take the sum.
As per the given question, we have to find the factor of the given trinomial. And the given trinomial to be factored is \[6{{x}^{2}}+13x+5\].
Here, in the given trinomial expression, we have both the b term and the c term positive. Then, on comparing the given trinomial expression \[6{{x}^{2}}+13x+5\] with the general form of trinomial expression \[a{{x}^{2}}+bx+c\], we have \[a=6\], \[b=13\] and \[c=5\].
Now, product of a and c terms, \[ac=30\] and \[b=13\]. Let the two numbers x and y, such that their sum, \[x+y=13\] and their product, \[xy=30\].
If the product of two numbers is 30, then the possible combination of numbers can be
\[\Rightarrow 1\times 30\] or \[3\times 10\] or \[5\times 6\].
We have to select the two numbers from the above combinations such that their sum is 13. For the above combination of numbers, their sum will be
\[\Rightarrow (1+30)=31\], \[(3+10)=13\] and \[(5+6)=11\].
From above, we can say that the two numbers are 3 and 10.
Then, we use this pair to split the middle term and factor by grouping:
\[\begin{align}
& \Rightarrow 6{{x}^{2}}+13x+5 \\
& \Rightarrow (6{{x}^{2}}+10x)+(3x+5) \\
& \Rightarrow 2x(3x+5)+1(3x+5) \\
& \Rightarrow (2x+1)(3x+5) \\
\end{align}\]
\[\therefore (2x+1)(3x+5)\] is the factor of the given trinomial \[6{{x}^{2}}+13x+5\].
Note:
If the last number that is c term is negative, then we will have to work with the difference of the two numbers which are factors. While solving such types of problems, there is a scope of making mistakes when we try to make a factor common. Calculation mistakes should be avoided to get the correct answer.
A trinomial is a polynomial with only three terms. If \[a{{x}^{2}}+bx+c\] is a trinomial, when it does not factor into a square, we factorize it by extension of factoring. That is, we use the sum-product method to factor the trinomial. We can check the factor trinomial by the fact that when simplified it is the same as the actual trinomial.
Complete step by step answer:
For a trinomial expression like \[a{{x}^{2}}+bx+c\], we try to write the c term as the product of two numbers, and b term as the sum (or difference) of the same two numbers. In the case of both being positive, it’s not hard to take the sum.
As per the given question, we have to find the factor of the given trinomial. And the given trinomial to be factored is \[6{{x}^{2}}+13x+5\].
Here, in the given trinomial expression, we have both the b term and the c term positive. Then, on comparing the given trinomial expression \[6{{x}^{2}}+13x+5\] with the general form of trinomial expression \[a{{x}^{2}}+bx+c\], we have \[a=6\], \[b=13\] and \[c=5\].
Now, product of a and c terms, \[ac=30\] and \[b=13\]. Let the two numbers x and y, such that their sum, \[x+y=13\] and their product, \[xy=30\].
If the product of two numbers is 30, then the possible combination of numbers can be
\[\Rightarrow 1\times 30\] or \[3\times 10\] or \[5\times 6\].
We have to select the two numbers from the above combinations such that their sum is 13. For the above combination of numbers, their sum will be
\[\Rightarrow (1+30)=31\], \[(3+10)=13\] and \[(5+6)=11\].
From above, we can say that the two numbers are 3 and 10.
Then, we use this pair to split the middle term and factor by grouping:
\[\begin{align}
& \Rightarrow 6{{x}^{2}}+13x+5 \\
& \Rightarrow (6{{x}^{2}}+10x)+(3x+5) \\
& \Rightarrow 2x(3x+5)+1(3x+5) \\
& \Rightarrow (2x+1)(3x+5) \\
\end{align}\]
\[\therefore (2x+1)(3x+5)\] is the factor of the given trinomial \[6{{x}^{2}}+13x+5\].
Note:
If the last number that is c term is negative, then we will have to work with the difference of the two numbers which are factors. While solving such types of problems, there is a scope of making mistakes when we try to make a factor common. Calculation mistakes should be avoided to get the correct answer.
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