Question

How do you factor \[{x^2} + 25x + 150\] ?

Answer 1

Hint: It is a simple quadratic equation in one variable \[x\]. We can easily solve this equation. There are two methods to find the factors of the given quadratic equation. Either we can use the factorization method or we can use the quadratic formula method to find the factors. Since it is a quadratic equation, it has two factors.

Complete step by step solution:
We can solve the given quadratic equation by using the formula method. the formula to solve an equation of the form \[a{x^2} + bx + c = 0\] is given by
 \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] (it is also called as Sreedhar acharya formula)
The first term is, \[{x^2}\] its coefficient is \[1\] .so the value of \[a = 1\]
The middle term is, \[ + 25x\] coefficient of \[x\] is \[ + 25\] . so, the value of \[b = 25\]
The last term is "the constant", so the value of \[c = 150\]
On substitution in formula, we get
\[x = \dfrac{{ - 25 \pm \sqrt {{{\left( {25} \right)}^2} - 4\left( 1 \right)\left( {150} \right)} }}{{2\left( 1 \right)}}\]
On simplification we get
 \[ \Rightarrow x = \dfrac{{ - 25 \pm \sqrt {\left( {625} \right) - \left( {600} \right)} }}{2}\]
 \[ \Rightarrow x = \dfrac{{ - 25 \pm \sqrt {25} }}{2}\]
 \[ \Rightarrow x = \dfrac{{ - 25 \pm 5}}{2}\]
Thus we have two roots,
 \[ \Rightarrow x = \dfrac{{ - 25 + 5}}{2}\] (or) \[x = \dfrac{{ - 25 - 5}}{2}\]
 on simplification we get
 \[ \Rightarrow x = \dfrac{{ - 20}}{2}\] (or) \[x = \dfrac{{ - 30}}{2}\]
 \[ \Rightarrow x = - 10\] (or) \[x = - 15\]
Therefore, the required factors are \[x = - 10\] & \[x = - 15\].

Note: Factors are whole numbers that are multiplied together to produce another number or we can also define factors as one of two or more numbers that divides a given number without a remainder. The original numbers are factors of the product number.to find the factors of a quadratic equation formula method is more useful than factorization method because formula methods always work to give the solution even if the solutions are not real numbers. And if we solve the above problem using the factorization method by taking the factors as 10 and 15, we will get the same solution.