
How do you factorize $7{x^2} - 28$?
Answer
542.1k+ views
Hint: Factorization formula for a quadratic polynomial –
A quadratic is a polynomial that is written like “$a{x^2} + bx + c$”. For an easy case of factoring, you can identify the two numbers that will not only multiply to equal the constant term “$c$” but also add up to equal “ $b$” the coefficient on the x-term.
Complete step by step answer:
In such questions first, we need to determine the common factor, which in this particular question is $7$ considering the given terms.
$7({x^2} - 4) \to (1)$
${x^2} - 4$ is a difference of squares which is factored in general as follows-
${a^2} - {b^2} = (a - b)(a + b)$
Here $a = x$ and $b = 2$
$ \Rightarrow {x^2} - 4 = (x - 2)(x + 2)$
Now, going back to the first equation-
$7{x^2} - 28 = 7(x - 2)(x + 2)$
Additional Information: In number theory, integer factorization is the decomposition of a composite number into a product of small integers. If these factors are further restricted to prime numbers, the process is called prime factorization.
The method of prime factorization is used to “break down” or express a given number as a product of prime numbers. More so, if a prime number occurs more than once in the factorization, it is usually expressed in exponential form to make it look more compact.
Note: Factorising formulas algebra is especially important when solving quadratic polynomials. When reducing formulas, we normally have to remove all the brackets, but in particular cases, for example with fractional formulas, sometimes we can use factorization to shorten a formula.
A quadratic is a polynomial that is written like “$a{x^2} + bx + c$”. For an easy case of factoring, you can identify the two numbers that will not only multiply to equal the constant term “$c$” but also add up to equal “ $b$” the coefficient on the x-term.
Complete step by step answer:
In such questions first, we need to determine the common factor, which in this particular question is $7$ considering the given terms.
$7({x^2} - 4) \to (1)$
${x^2} - 4$ is a difference of squares which is factored in general as follows-
${a^2} - {b^2} = (a - b)(a + b)$
Here $a = x$ and $b = 2$
$ \Rightarrow {x^2} - 4 = (x - 2)(x + 2)$
Now, going back to the first equation-
$7{x^2} - 28 = 7(x - 2)(x + 2)$
Additional Information: In number theory, integer factorization is the decomposition of a composite number into a product of small integers. If these factors are further restricted to prime numbers, the process is called prime factorization.
The method of prime factorization is used to “break down” or express a given number as a product of prime numbers. More so, if a prime number occurs more than once in the factorization, it is usually expressed in exponential form to make it look more compact.
Note: Factorising formulas algebra is especially important when solving quadratic polynomials. When reducing formulas, we normally have to remove all the brackets, but in particular cases, for example with fractional formulas, sometimes we can use factorization to shorten a formula.
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