Question

Factor \[49{{x}^{2}}-16?\]

Answer 1

Hint: Here a quadratic equation is given which we have to solve.
For solving the given quadratic equation we have to use the factorization method.
The quadratic equation is expressed in the form of, \[a{{x}^{2}}+bx+c=0\]
Factorization method is nothing but breaking the large number of polynomials into small factors.
It always gives two answers.
The simplest way to factorize the given number is to multiply such two numbers whose answer will be constant \['c'\] and the addition or subtraction will result in \['b'\].

Complete step by step solution:
Given that, there is an equation
\[49{{x}^{2}}-16\,\,\,\,\,.........(i)\]
And we have made its factors.
As we know that, the square root of\[49\]is \[7\] i.e. \[\sqrt{49}=\pm 7\]and square root of \[16\]is \[4\]i.e.
\[\sqrt{16}=\pm 4\]
So, we can write the equation \[(i)\]as,
\[(7{{x}^{2}})-{{\left( 4 \right)}^{2}}\,\,\,\,\,\,......(ii)\]
As both the terms are perfect squares.
So here we can use the formula of difference of square, which is given as,
\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\,\,\,\,\,...........(iii)\]
Compare the equation \[(ii)\] with equation \[(iii)\] we can say that,
\[a=7x\] and \[b=4\]

\[\therefore \] The equation \[(ii)\] can become.
\[{{\left( 7x \right)}^{2}}-{{\left( 4 \right)}^{2}}=\left( 7x+4 \right)\left( 7x-4 \right)\]
The above equation can be rewrite as, \[\left( 7x+4 \right)\left( 7x-4 \right)=0\]
\[\therefore \,\,\,\,7x+4=0\]and \[7x-4=0\]
\[\therefore \,\,\,\,7x=-4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7x=4\]
\[x=-\dfrac{4}{7}\] and \[x=\dfrac{4}{7}\]
Therefore the factors of given equation is,
\[x=\left\{ -\dfrac{4}{7},\dfrac{4}{7} \right\}\]

Additional information:
The factorization method is nothing but breaking the given number or matrix or polynomial into two small factors. Which means that the product of those two factors should give original value.
The quadratic equation is written in the form of
\[a{{x}^{2}}+bx+c=0\]
Where \[a,b,c\] are the numbers.
For easy factoring we have to find such two numbers which will not only equal to the constant \['c'\] after multiplication but also its addition will be equal to \['b'\]
In general, the factoring is important to remove all brackets from the equation.
The term is that which has to be added or subtracted while performing factorization. And the factors is that which has to be multiplied
\[\therefore \] Sum \[=\] Term \[+\]Term
Product \[=\] Factor \[\times \]Factor

Note: In this question, a quadratic equation is given which we have to solve.
For that we have to simplify it using the factorization method.
As we know, factorization means breaking large numbers or polynomials into small one.
In factorization we always get two answers.
In this question, the equation is, \[49{{x}^{2}}-16,\] If we use the difference of square formula we can easily get the answer because both the terms that \[49\] and \[16\] have perfect squares. So instead of using standard formula of quadratic equation, we are using the formula of difference square, and we get the answer as, \[x=\left\{ -\dfrac{4}{7},\dfrac{4}{7} \right\}.\]