Question

How do you simplify $ \dfrac{{2{z^2} - 11z + 15}}{{{z^2} - 9}} $ ?

Answer 1

Hint: A rational number is any number of the form ab, where b≠0. A rational expression is any algebraic expression of the form a(x)b(x), where b≠0.Replacing the number in the numerator and denominator with the expression makes it a rational expression. Also the values of the variable that causes the denominator of a rational expression to be zero is referred to as restrictions and must be excluded from the set of possible values for the variable whereas in our given question it is square in our denominator .

Complete step-by-step answer:
At First we will factor the numerator $ 2{z^2} - 11z + 15 $ and we get,
Given a quadratic equation, let it be $ f(b) $
 $ f(z) = 2{z^2} - 11z + 15 $
Comparing the equation with the standard Quadratic equation $ a{x^2} + bx + c $
a becomes 2
b becomes -11
And c becomes 15
To find the quadratic factorization we’ll use splitting up the middle term method
So first calculate the product of coefficient of $ {z^2} $ and the constant term which comes to be
 $ = 2 \times 15 = 30 $
Now the second Step is to find the 2 factors of the number 2 such that the whether addition or subtraction of those numbers is equal to the middle term or coefficient of x and the product of those factors results in the value of constant .
So if we factorize -11 ,the answer comes to be -6 and -5 as $ - 6 - 5 = - 11 $ that is the middle term . and $ - 6 \times ( - 5) = 30 $ which is perfectly equal to the constant value.
Now writing the middle term sum of the factors obtained ,so equation $ f(z) $ becomes
 $ f(z) = 2{z^2} - 6z - 5z + 15 $
Now taking common from the first 2 terms and last 2 terms
 $ f(z) = 2z(z - 3) - 5(z - 3) $
Finding the common binomial parenthesis, the equation becomes
 $ f(z) = (2z - 5)(z - 3) $
Hence , We have successfully factorized our quadratic equation.
Therefore the factors are $ (2z - 5) $ and $ (z - 3) $
So, now we are going to rewrite the rational expression as the factors we got as to create the simplified expression and cancel out the common factor we will get as-
 $ \dfrac{{2{z^2} - 11z + 15}}{{{z^2} - 9}} $
Also by applying the the identity in the denominator as follows = \[{a^2} - {b^2} = (a + b)(a - b)\]
 $\Rightarrow \dfrac{{2z(z - 3) - 5(z - 3)}}{{(z - 3)(z + 3)}} $ , here we are cancelling $ (z - 3) $ as a common factor which will give the result as
 $ \dfrac{{2z - 5}}{{z + 3}} $ .
This is our final answer $ \dfrac{{2z - 5}}{{z + 3}} $ .
So, the correct answer is “ $ \dfrac{{2z - 5}}{{z + 3}} $ ”.

Note: Remember that you cannot “cancel out” common factors until both the numerator and denominator have been factored .
The restriction is $ z \ne - 3 $ because the value for z would have made the denominator of the original rational expression to zero.
The restrictions to the domain of a rational expression are determined by the denominator. Ignore the numerator when finding those restrictions.
The resulting rational expression is equivalent if it shares the same domain. Therefore, we must make note of the restrictions and write.
 Rational expressions are not given in factored form. If this is the case, factor first and then cancel.