Question

How do you simplify $\dfrac{{25 - {x^2}}}{{{x^2} - 10x + 25}}$?

Answer 1

Hint: First, we have to simplify the numerator by rewriting $25$ as ${5^2}$ and then using the difference of squares formula to factor it. Next, simplify the denominator by factoring using the perfect square rule. Next, take $ - 1$ common from the numerator and cancel the common factors. We will get the desired result.

Formula used:
Difference of squares formula: ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$
Perfect square trinomial rule: ${a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}$

Complete step by step answer:
We have to simplify $\dfrac{{25 - {x^2}}}{{{x^2} - 10x + 25}}$.
First, we will simplify the numerator.
Rewrite $25$ as ${5^2}$.
$\dfrac{{{5^2} - {x^2}}}{{{x^2} - 10x + 25}}$
Since both terms are perfect squares, factor using the difference of squares formula, ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ where $a = 5$ and $b = x$.
$\dfrac{{\left( {5 - x} \right)\left( {5 + x} \right)}}{{{x^2} - 10x + 25}}$
Now, simplify the denominator.
Factor using the perfect square rule.
Rewrite $25$ as ${5^2}$.
$\dfrac{{\left( {5 - x} \right)\left( {5 + x} \right)}}{{{x^2} - 10x + {5^2}}}$
Check the middle term by multiplying $2ab$ and compare this result with the middle term in the original expression.
$2ab = 2 \cdot x \cdot 5$
Simplify.
$2ab = 10x$
Factor using the perfect square trinomial rule ${a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}$ where $a = x$ and $b = 5$.
$\dfrac{{\left( {5 - x} \right)\left( {5 + x} \right)}}{{{{\left( {x - 5} \right)}^2}}}$
Take $ - 1$ common from the numerator.
$ \Rightarrow - \dfrac{{\left( {x - 5} \right)\left( {x + 5} \right)}}{{{{\left( {x - 5} \right)}^2}}}$
Cancel the common factors.
Factor $x - 5$ out of ${\left( {x - 5} \right)^2}$.
$ \Rightarrow - \dfrac{{\left( {x - 5} \right)\left( {x + 5} \right)}}{{\left( {x - 5} \right)\left( {x - 5} \right)}}$
Cancel the common factor.
$ \Rightarrow - \dfrac{{\left( {x + 5} \right)}}{{\left( {x - 5} \right)}}$

Therefore, $\dfrac{{25 - {x^2}}}{{{x^2} - 10x + 25}} = - \dfrac{{x + 5}}{{x - 5}}$

Note: We can also factorize trinomial ${x^2} - 10x + 25$ by splitting the middle term.
For factorising an algebraic expression of the type $a{x^2} + bx + c$, we find two factors $p$ and $q$ such that
$ac = pq$ and $p + q = b$
Step by step solution:
Given, ${x^2} - 10x + 25$
We have to factor this trinomial.
To factor this trinomial, first we have to find the product of the first and last constant term of the expression.
Here, the first constant term in ${x^2} - 10x + 25$ is $1$, as it is the coefficient of ${x^2}$ and last constant term is $25$, as it is a constant value.
Now, we have to multiply the coefficient of ${x^2}$ with the constant value in ${x^2} - 10x + 25$, i.e., multiply $1$ with $25$.
Multiplying $1$ and $25$, we get
$1 \times 25 = 25$
Now, we have to find the factors of $25$ in such a way that addition or subtraction of those factors is the middle constant term.
Middle constant term or coefficient of $x$ in ${x^2} - 10x + 25$ is $ - 10$.
So, we have to find two factors of $25$, which on multiplying gives $25$ and in addition gives $ - 10$.
We can do this by determining all factors of $25$.
Factors of $25$ are $ \pm 1, \pm 5, \pm 25$.
Now among these values find two factors of $25$, which on multiplying gives $25$ and in addition gives $ - 10$.
After observing, we can see that
$\left( { - 5} \right) \times \left( { - 5} \right) = 25$ and $\left( { - 5} \right) + \left( { - 5} \right) = - 10$
So, these factors are suitable for factorising the given trinomial.
Now, the next step is to split the middle constant term or coefficient of $x$ in these factors.
That is, write $ - 10x$ as $ - 5x - 5x$ in ${x^2} - 10x + 25$.
After writing $ - 10x$ as $ - 5x - 5x$ in ${x^2} - 10x + 25$, we get
$ \Rightarrow {x^2} - 5x - 5x + 25$
Now, taking $x$ common in ${x^2} - 5x$ and putting in above equation, we get
$ \Rightarrow x\left( {x - 5} \right) - 5x + 25$
Now, taking $\left( { - 5} \right)$ common in $ - 5x + 25$ and putting in above equation, we get
$ \Rightarrow x\left( {x - 5} \right) - 5\left( {x - 5} \right)$
Now, taking \[\left( {x - 5} \right)\] common in \[x\left( {x - 5} \right) - 5\left( {x - 5} \right)\] and putting in above equation, we get
$ \Rightarrow \left( {x - 5} \right)\left( {x - 5} \right)$
Therefore, the trinomial ${x^2} - 10x + 25$ can be factored as $\left( {x - 5} \right)\left( {x - 5} \right)$.