Question

How do you multiply \[\left( {8x - 11} \right)\left( {8x + 11} \right)\]?

Answer 1

Hint: In order to multiply the above given expression , use the identity $\left( {A - B} \right)\left( {A + B} \right) = {A^2} - {B^2}$ to rewrite the given expression by considering A as $8x$ and B as $11$ and simplify the expression to get the required result.

Complete step-by-step solution:
We are given a polynomial having one variable $x$ in the term.
Let’s suppose the function given be $f\left( y \right)$
$\Rightarrow f\left( x \right) = \left( {8x - 11} \right)\left( {8x + 11} \right)$
In order to multiply the above binomial expression, we will be using the identity $\left( {A - B} \right)\left( {A + B} \right) = {A^2} - {B^2}$ to rewrite the given expression by considering A as $8x$ and B as $11$.
Our expression now becomes,
$\Rightarrow f\left( x \right) = {\left( {8x} \right)^2} - {\left( {11} \right)^2}$
Simplifying the above expression, we get a quadratic equation as
$\Rightarrow f\left( x \right) = 64{x^2} - 121$

Therefore, the multiplied form of the given expression is $64{x^2} - 121$.

Additional Information:
Quadratic Equation: A quadratic equation is an equation which can be represented in the form of $a{x^2} + bx + c$ where $x$ is the unknown variable and a,b,c are the numbers known where $a \ne 0$. If $a = 0$ then the equation will become a linear equation and will no longer be quadratic .
The degree of the quadratic equation is of the order 2.
In order to determine the roots to a quadratic equation, there are couple of ways,
1.Using splitting up the middle term method:
let $a{x^2} + bx + c$
 calculate the product of coefficient of ${x^2}$ and the constant term and factorise it into two factors in a way that either addition or subtraction of the two gives the middle term and multiplication gives the product value.
2.You can also alternatively use a direct method which uses Quadratic Formula to find both roots of a quadratic equation as
$x1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and $x2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
$x_1,x_2$ are root to quadratic equation $a{x^2} + bx + c$

Note:
You can alternatively get the required result by simply multiplying the terms using the FOIL rule which means First, Outside,Inside,Last.
$
\Rightarrow f\left( x \right) = \left( {8x - 11} \right)\left( {8x + 11} \right) \\
\Rightarrow 8x\left( {8x + 11} \right) - 11\left( {8x + 11} \right) \\
 \Rightarrow 64{x^2} + 88x - 88x - 121 \\
 \Rightarrow 64{x^2} - 121 \\
 $