Answer
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Hint: Before attempting this question, one should have prior knowledge about the concept of factorization and also remember to use splitting the middle term method and suppose (x – p) be another factor of a given quadratic equation, using this information can help you to approach the solution of the problem.
Complete step-by-step answer:
According to the given information we have a quadratic equation i.e. ${x^2} + x - 20$ and (x + 5)
Let (x – p) be another factor of the given quadratic equation ${x^2} + x - 20$
Therefore, $\left( {x + 5} \right)\left( {x-p} \right) = {x^2} + x - 20$
Now using the splitting, the middle term method for the given quadratic equation ${x^2} + x - 20$
$\left( {x + 5} \right)\left( {x-p} \right) = {x^2} + (5 - 4)x - 20$
$ \Rightarrow \left( {x + 5} \right)\left( {x-p} \right) = {x^2} + 5x - 4x - 20$
$ \Rightarrow \left( {x + 5} \right)\left( {x-p} \right) = x\left( {x + 5} \right) - 4\left( {x + 5} \right)$
$ \Rightarrow \left( {x + 5} \right)\left( {x-p} \right) = \left( {x - 4} \right)\left( {x + 5} \right)$
$ \Rightarrow $x – p = x – 4
$ \Rightarrow $x – x = p – 4
$ \Rightarrow $0 = p – 4
$ \Rightarrow $p = 4
Substituting, the value of p in another factor of the given quadratic equation we get
(x – 4)
Therefore, another factor is (x – 4)
Hence, option A is the correct option.
Note: In the above problem we came across the term “quadratic equation” which can be explained as the polynomial equation formed by many terms like numbers and variables in the polynomial equation the variable with the highest power defines the numbers of roots the equation can have, for example, if we have a polynomial equation ${x^2} - 2x - 35$ as you can see that this equation has variable x which has the highest power of 2, therefore, this equation will have 2 roots.
Complete step-by-step answer:
According to the given information we have a quadratic equation i.e. ${x^2} + x - 20$ and (x + 5)
Let (x – p) be another factor of the given quadratic equation ${x^2} + x - 20$
Therefore, $\left( {x + 5} \right)\left( {x-p} \right) = {x^2} + x - 20$
Now using the splitting, the middle term method for the given quadratic equation ${x^2} + x - 20$
$\left( {x + 5} \right)\left( {x-p} \right) = {x^2} + (5 - 4)x - 20$
$ \Rightarrow \left( {x + 5} \right)\left( {x-p} \right) = {x^2} + 5x - 4x - 20$
$ \Rightarrow \left( {x + 5} \right)\left( {x-p} \right) = x\left( {x + 5} \right) - 4\left( {x + 5} \right)$
$ \Rightarrow \left( {x + 5} \right)\left( {x-p} \right) = \left( {x - 4} \right)\left( {x + 5} \right)$
$ \Rightarrow $x – p = x – 4
$ \Rightarrow $x – x = p – 4
$ \Rightarrow $0 = p – 4
$ \Rightarrow $p = 4
Substituting, the value of p in another factor of the given quadratic equation we get
(x – 4)
Therefore, another factor is (x – 4)
Hence, option A is the correct option.
Note: In the above problem we came across the term “quadratic equation” which can be explained as the polynomial equation formed by many terms like numbers and variables in the polynomial equation the variable with the highest power defines the numbers of roots the equation can have, for example, if we have a polynomial equation ${x^2} - 2x - 35$ as you can see that this equation has variable x which has the highest power of 2, therefore, this equation will have 2 roots.
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