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# What is the value of the constant term in the expansion given as $23{{x}^{3}}+12{{x}^{2}}-6x-12$?(a) 12(b) 6(c) -6(d) -12

Last updated date: 20th Jun 2024
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Hint: We start solving the problem by assigning a function to the given expansion. We use the fact that we get the constant term by substituting ‘0’ in place of ‘x’ in the expansion. So, we substitute ‘0’ in place of ‘x’ to get the required value of the constant term.

Complete step-by-step solution:
Given that we have an expansion $23{{x}^{3}}+12{{x}^{2}}-6x-12$, and we need to find the value of the constant term in that given expansion.
Let us represent the given expansion with f(x). So, we have $f(x)=23{{x}^{3}}+12{{x}^{2}}-6x-12$ ---(1).
We can see that the given expansion is polynomial of degree 3 (we know that degree is the maximum power of x in a given polynomial of x). We know that to find the value of the constant term in expansion, we substitute x = 0 in the expansion.
So, let us substitute x = 0 in the polynomial f(x).
So, we have got $f\left( 0 \right)=23{{\left( 0 \right)}^{3}}+12{{\left( 0 \right)}^{2}}-6\left( 0 \right)-12$.
We have got $f\left( 0 \right)=23\left( 0 \right)+12\left( 0 \right)-6\left( 0 \right)-12$.
We have got f(0) = $0 + 0 – 0 – 12$.
We have got f(0) = $–12$.
So, we have got the value of f(0) as $–12$.
Since f(0) is the value of the constant term of the expansion, We get the value of the constant term of the expansion as $–12$.
$\therefore$ The value of the constant term in the expansion $23{{x}^{3}}+12{{x}^{2}}-6x-12$ is –12.

Note: We solved this problem by substituting ‘0’ in place of ‘x’, this method can also be adopted in the cases where ‘x’ contains powers in the fractions. Similarly, we can expect problems to tell the value of co-efficient of ‘x’ and other higher powers of ‘x’.