Question

How do you solve $2{p^3} - 3{p^2} - 98p + 147 = 0?$

Answer 1

Hint: You have given a polynomial equation of degree three, and you can solve this equation by factoring the left hand side expression. If seen properly then one could see that the given equation can be factored by grouping first and third term, and second and fourth term. And then after factoring the equation, equate each term with zero separately to get the required set of solutions for the given polynomial equation.

Formula used: Algebraic identity of difference of two square terms: ${a^2} - {b^2} = (a + b)(a - b)$


Complete step by step solution:
 In order to solve the given equation, that is $2{p^3} - 3{p^2} - 98p + 147 = 0$ we will factorize the expression present at the left hand side of the equation as follows
On observing the equation carefully, we can see that if we group the terms with degrees three and one and group the terms with degrees two and zero, then the equation will be easily factorized,
So,
$
   \Rightarrow 2{p^3} - 3{p^2} - 98p + 147 = 0 \\
   \Rightarrow \left( {2{p^3} - 98p} \right) + \left( { - 3{p^2} + 147} \right) = 0 \\
 $
Now, taking $2p$ common in the first group and $ - 3$ common in the second group, we will get
$ \Rightarrow 2p\left( {{p^2} - 49} \right) - 3\left( {{p^2} - 49} \right) = 0$
By taking $\left( {{p^2} - 49} \right)$ common in the above equation, we can write it as
$ \Rightarrow \left( {{p^2} - 49} \right)\left( {2p - 3} \right) = 0$
We can also write it as
$ \Rightarrow \left( {{p^2} - {7^2}} \right)\left( {2p - 3} \right) = 0$
Using the algebraic identity ${a^2} - {b^2} = (a + b)(a - b)$ to simplify further, we will get
$ \Rightarrow \left( {p + 7} \right)\left( {p - 7} \right)\left( {2p - 3} \right) = 0$
Now, equating each term with zero to get the solution,
$
   \Rightarrow \left( {p + 7} \right) = 0,\;\left( {p - 7} \right) = 0,\;{\text{and}}\;\left( {2p - 3} \right) = 0 \\
   \Rightarrow p = - 7,\;p = 7,\;{\text{and}}\;2p = 3 \\
   \Rightarrow p = - 7,\;p = 7,\;{\text{and}}\;p = \dfrac{3}{2} \\
 $

Therefore $p = - 7,\;7,\;{\text{and}}\;\dfrac{3}{2}$ is the required solution.

Note: There is one more alternative method to solve this equation, first by hit and trial and then after getting one factor of the equation, divide the equation with that factor to get another two factors and eventually the solution. But sometimes this method takes much time because generally we try with small integers and equations having some greater integer or only fraction will create trouble in this method.