Question

Find the zeros by rewriting the function $h(x) = {x^2} - 4x - 77$ in intercept form?

Answer 1

Hint: This problem deals with factorizing the given expression and then finding the zeros of the given function. This can be done either by the method of completing the square or just factoring and solving the quadratic equation. To solve $a{x^2} + bx + c$, expression in $x$, by factoring the given expression. But here we are adding and subtracting some terms in order to factor.

Complete step by step solution:
Given the quadratic expression is $h(x) = {x^2} - 4x - 77$, consider it as given below:
$ \Rightarrow h(x) = {x^2} - 4x - 77$
Now expressing the above expression such that the $x$ term is split into the factors of the product of the ${x^2}$ term and the constant term which is equal to 77, now the factors of 77, which the difference of the factors is equal to the coefficient of the $x$ term are 7 and 11, where these product is also equal to 77, hence the term $ - 4x$ is expressed as the difference of $7x$ and $11x$, as shown below:
$ \Rightarrow h(x) = {x^2} - 4x - 77$
$ \Rightarrow h(x) = {x^2} + 7x - 11x - 77$

Now taking the term $x$ common from the first two terms, and taking the number 11 common from the second two terms, which is shown below:
$ \Rightarrow h(x) = x\left( {x + 7} \right) - 11\left( {x + 7} \right)$

Now taking the term $\left( {x + 7} \right)$ common in the above expression, as shown below:
$ \Rightarrow h(x) = \left( {x - 11} \right)\left( {x + 7} \right)$
So here we factorized the given quadratic expression into two factors, which is shown below:
$ \Rightarrow {x^2} - 4x - 77 = \left( {x - 11} \right)\left( {x + 7} \right)$
We get the zeros of the expressions by equating the factors to zero.

The zeros of ${x^2} - 4x - 77$ are \[11\] and \[ - 7\].

Note: Instead of first factoring and then solving for $x$, we can directly say the value of $x$ from the given equation is directly equated to zero, as ${x^2} - 4x - 77 = 0$, then after finding the roots, and then factorize.