Question

If HCF & LCM of two quadratic expressions are \[\left( {x - 5} \right)\] and \[{x^3} - 19x - 30\] then find the two expressions.
A. \[p\left( x \right) = \left( {x - 5} \right)\left( {x + 2} \right)\] and \[q\left( x \right) = \left( {x - 5} \right)\left( {x + 3} \right)\]
B. \[p\left( x \right) = \left( {x - 2} \right)\left( {x - 8} \right)\] and \[q\left( x \right) = \left( {x - 1} \right)\left( {x - 3} \right)\]
C. \[p\left( x \right) = \left( {x + 1} \right)\left( {x - 2} \right)\] and \[q\left( x \right) = \left( {x - 1} \right)\left( {x - 3} \right)\]
D. \[p\left( x \right) = \left( {x - 1} \right)\left( {x - 2} \right)\] and \[q\left( x \right) = \left( {x - 1} \right)\left( {x - 8} \right)\]

Answer 1

Hint:
In order to solve this question, first of all, we will find the factors of the cubic polynomial which is the LCM. We will find the factors with the help of a long division method. After finding them, we will see that one factor will be common in both HCF as well as LCM, that factor will be present in both the required expressions, and the other two factors will be divided among the two expressions. We will verify this by the formula that the product of two quadratic expressions is equal to the product of their LCM and HCF. Hence, we will reach our required answer.

Complete step by step solution:
Let the two quadratic expressions be \[p\left( x \right)\] and \[q\left( x \right)\].
It is given that their HCF is \[\left( {x - 5} \right)\] and, LCM is \[{x^3} - 19x - 30\].
Now, by hit and try method we will try to find a factor of this cubic polynomial.
We will substitute \[x = 0,1, - 1, - 2\] to find the root of \[{x^3} - 19x - 30\].
Substituting \[x = - 2\] in the equation \[{x^3} - 19x - 30\], then,
\[{x^3} - 19x - 30 = {\left( { - 2} \right)^3} - 19\left( { - 2} \right) - 30\]
Applying the exponent on the terms, we get
\[ \Rightarrow {x^3} - 19x - 30 = - 8 + 38 - 30\]
Adding the terms, we get
\[ \Rightarrow {x^3} - 19x - 30 = 38 - 38 = 0\]
So, when \[x = - 2\] the equation is equal to 0.
Hence, \[\left( {x + 2} \right)\] is a factor of this cubic polynomial.
As we know, Dividend \[=\] Divisor \[ \times \] Quotient + Remainder
Hence,
\[{x^3} - 19x - 30{\rm{ }} = \left( {x + 2} \right) \times \left( {{x^2} - 2x - 15} \right) + 0\] …………………………………..\[\left( 1 \right)\]
Now, first of all we will solve the quadratic equation i.e the Quotient using middle term splitting,
Hence,
\[{x^2} - 2x - 15\] can be written as :
\[{x^2} - 5x + 3x - 15\]
(Middle term splitting is a method in which we divide the middle term in two factors such that their product equals to the product of the first and the last term.)
Hence, making two factors by taking common we get,
\[{x^2} - 2x - 15 = x\left( {x - 5} \right) + 3\left( {x - 5} \right)\]
Now, taking the same brackets as one common bracket, we get,
\[{x^2} - 2x - 15 = \left( {x - 5} \right)\left( {x + 3} \right)\]
Hence, substituting this value in equation \[\left( 1 \right)\] , we get,
\[{x^3} - 19x - 30{\rm{ }} = \left( {x + 2} \right)\left( {x - 5} \right)\left( {x + 3} \right)\]
Now, HCF \[ = \left( {x - 5} \right)\]
And LCM \[ = \left( {x + 2} \right)\left( {x - 5} \right)\left( {x + 3} \right)\]
Clearly,
 \[\left( {x - 5} \right)\] is common
Hence, it will be there in both the quadratic expressions.
Now,
First expression will be
\[\left( {x - 5} \right)\left( {x + 2} \right)\]
And, second expression will be
\[\left( {x - 5} \right)\left( {x + 3} \right)\]
This is because of the fact that when we multiply two quadratic expressions, then,
\[p\left( x \right) \times q\left( x \right) = \]HCF \[ \times \] LCM
Hence, clearly,
\[\left( {x - 5} \right)\left( {x + 2} \right)\left( {x - 5} \right)\left( {x + 3} \right) = \left( {x - 5} \right)\left( {x + 2} \right)\left( {x - 5} \right)\left( {x + 3} \right)\]
Hence, LHS is equal to RHS.
Therefore, the required two quadratic expressions are:
\[p\left( x \right) = \left( {x - 5} \right)\left( {x + 2} \right)\] and \[q\left( x \right) = \left( {x - 5} \right)\left( {x + 3} \right)\]

Hence, option A is the correct answer.

Note:
We should know how to do long division methods in order to solve this question.This question required a high level of attention as we could go wrong in so many steps. For example, while doing the long division method, we could forget to change the signs and hence, we will never reach towards the remainder \[0\] . Also, in the end when we were doing middle term splitting, by mistake we could interchange the signs and make our solution wrong. Hence, these were all the main areas where we should have paid attention to make our solution correct.