Question

How do you find the factors of \[2{{x}^{2}}+11x+15\]?

Answer 1

Hint: This type of problem is based on the concept of factorisation. First, we have to consider the given equation and split the middle term of the equation in such a way that it has a common term in both the first and last term. Here, the middle term is 11x, so we can write 11x as the sum of 5x and 6x. Then, take the common terms from the obtained equation and represent the equation as a product of two functions of x. Thus, the two functions are the factors of the given equation which is the required answer.

Complete step by step answer:
According to the question, we are asked to find the factors of \[2{{x}^{2}}+11x+15\].
We have been given the quadratic equation is \[2{{x}^{2}}+11x+15\]. ---------(1)
Let us first divide the middle term in such a way that the sum of the two terms is equal to 11 and the product of the two terms is equal to 30.
We know that \[6\times 5=30\] and 6+5=11.
Let us substitute in equation (1).
We get
\[2{{x}^{2}}+11x+15=2{{x}^{2}}+\left( 6+5 \right)x+15\]
On using distributive property in the obtained equation, that is, \[a\left( b+c \right)=ab+ac\]
We get
\[2{{x}^{2}}+11x+15=2{{x}^{2}}+6x+5x+15\]
Let us now find the common term.
\[2{{x}^{2}}+6x+5x+15=2{{x}^{2}}+2\times 3x+5x+3\times 5\]
Here, 2x is the common term from the first two terms and 5 is common from the last two terms.
Therefore, \[2{{x}^{2}}+6x+5x+15=2x\left( x+3 \right)+5\left( x+3 \right)\].
From the obtained equation, find that (x+3) is common.
On taking out (x+3) common from the two terms, we get
\[2{{x}^{2}}+6x+5x+15=\left( x+3 \right)\left( 2x+5 \right)\]
Therefore, the factors are (x+3) and (2x+5).

Hence, the factors of the given equation \[2{{x}^{2}}+11x+15\] are (x+3) and (2x+5).

Note: Whenever we get such type of problems, we should make necessary calculations to the given quadratic equation and then take out the common terms out of the bracket to obtain the factors. Also, avoid calculation mistakes based on the sign conventions. Here, we can solve this question by using quadratic formula, that is, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. The factors will be \[\left( x-\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \right)\]. In this a=2, b=11 and c=15. Substitute these values in the given formula and we can find the factors.