Question

How do you factor completely: $4{x^3} + 12{x^2} + 3x + 9$?

Answer 1

Hint: Here we have to factorize the linear polynomial. Factorization is defined as the process in which we break a number or a polynomial into a product of many factors of other polynomials, which when multiplied gives the original number. We will factorize the polynomial by taking $4{x^2}$ as a common factor from the first two terms and $3$ as a common factor from the last two terms.

Complete step-by-step solution:
Factorization is defined as the process of expressing or decomposing a number or an algebraic expression as a product of its factors.
In the question we have to factorize the polynomial $4{x^3} + 12{x^2} + 3x + 9$.
Polynomial can be defined as the expression consisting of coefficients and variables which are also known as intermediates. These are generally a sum or difference of variables and exponents. Each part of the polynomial is known as a “term”.
So, we can factorize the polynomial $4{x^3} + 12{x^2} + 3x + 9$ by taking $4{x^2}$ as a common factor from the first two terms and $3$ as a common factor from the last two terms.
Therefore,
$ \Rightarrow 4{x^3} + 12{x^2} + 3x + 9 = 4{x^2}(x + 3) + 3(x + 3)$
$ \Rightarrow (4{x^2} + 3)(x + 3)$
The remaining quadratic factor $4{x^2} + 3$ cannot be factored into linear factors with real coefficients.

Hence, $(4{x^2} + 3)(x + 3)$ is a factor of the polynomial $4{x^3} + 12{x^2} + 3x + 9$.

Note: In factorization there is an important term ‘Factor’. The factor is defined as the numbers, algebraic variables or an algebraic expression which divides the number or an algebraic expression without leaving any remainder. For example- the factors of $21 = 3 \times 7$, the numbers $3,7$ are the factors of $21$ and divide it without leaving any remainder. Similarly, for algebraic expression the factor of algebraic expression $6abc = 2 \times 3 \times a \times b \times c$.