How do you factor $64{x^2} - 49$?

Hint: First take $64$ common from the given equation and then divide both sides of the equation by $64$. Next, compare the given quadratic equation to the standard quadratic equation and find the value of numbers $a$, $b$ and $c$ in the given equation. Then, substitute the values of $a$, $b$ and $c$ in the formula of discriminant and find the discriminant of the given equation. Finally, put the values of $a$, $b$ and $D$ in the roots of the quadratic equation formula and get the desired result.

Formula used:
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step solution:
First, equate this polynomial with zero and make it an equation.
$ \Rightarrow 64{x^2} - 49 = 0$
We know that an equation of the form $a{x^2} + bx + c = 0$, $a,b,c,x \in R$, is called a Real Quadratic Equation.
The numbers $a$, $b$ and $c$ are called the coefficients of the equation.
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
So, first we will take $64$ common from the given equation.
$ \Rightarrow 64\left( {{x^2} - \dfrac{{49}}{{64}}} \right) = 0$
Divide both sides of the equation by $64$.
$ \Rightarrow {x^2} - \dfrac{{49}}{{64}} = 0$
Next, compare ${x^2} - \dfrac{{49}}{{64}} = 0$ quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing ${x^2} - \dfrac{{49}}{{64}} = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 1$, $b = 0$ and $c = - \dfrac{{49}}{{64}}$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( 0 \right)^2} - 4\left( 1 \right)\left( { - \dfrac{{49}}{{64}}} \right)$
After simplifying the result, we get
$ \Rightarrow D = \dfrac{{49}}{{16}}$
Which means the given equation has real roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$ \Rightarrow x = \dfrac{{ - 0 \pm \dfrac{7}{4}}}{{2 \times 1}}$
Divide numerator and denominator by $2$, we get
$ \Rightarrow x = \pm \dfrac{7}{8}$
$ \Rightarrow 8x = 7$ and $8x = - 7$
$ \Rightarrow 8x - 7 = 0$ and $8x + 7 = 0$

Therefore, the trinomial $64{x^2} - 49$ can be factored as $\left( {8x - 7} \right)\left( {8x + 7} \right)$.

Note: We can also factorize a given trinomial using algebraic identity.
Algebraic identity: ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$
So, rewrite $64{x^2}$ as ${\left( {8x} \right)^2}$.
$ \Rightarrow {\left( {8x} \right)^2} - 49$
Now, rewrite $49$ as ${7^2}$.
$ \Rightarrow {\left( {8x} \right)^2} - {7^2}$
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 $a = 8x$ and $b = 7$.
$ \Rightarrow \left( {8x - 7} \right)\left( {8x + 7} \right)$
Final Solution: Therefore, the trinomial $64{x^2} - 49$ can be factored as $\left( {8x - 7} \right)\left( {8x + 7} \right)$.