Question

Multiply the given polynomial $\left( 2{{x}^{3}}-3{{x}^{2}}+2x-3 \right)$ by $\left( 3{{x}^{2}}-4x+1 \right)$
(a) $6{{x}^{5}}-17{{x}^{4}}+20{{x}^{3}}-20{{x}^{2}}+14x-3$
(b) $6{{x}^{5}}+17{{x}^{4}}+20{{x}^{3}}-20{{x}^{2}}+14x-3$
(c) $6{{x}^{5}}-17{{x}^{4}}-20{{x}^{3}}-20{{x}^{2}}+14x-3$
(d) None of these

Answer 1

Hint: To multiply the two given algebraic expressions, multiply each term of the first expression with all the terms of the second expression, and then add them. Use the law of exponents, which says that ${{x}^{a}}\times {{x}^{b}}={{x}^{a+b}}$. Simplify the terms with the same power of exponents to find the value of the product of two expressions.

Complete step by step answer:
We have to multiply two algebraic expressions $\left( 2{{x}^{3}}-3{{x}^{2}}+2x-3 \right)$ and $\left( 3{{x}^{2}}-4x+1 \right)$.
To do so, we will multiply each term of the first expression with all the terms of the second expression and then add them up.
Thus, we have $\left( 2{{x}^{3}}-3{{x}^{2}}+2x-3 \right)\left( 3{{x}^{2}}-4x+1 \right)=2{{x}^{3}}\left( 3{{x}^{2}}-4x+1 \right)-3{{x}^{2}}\left( 3{{x}^{2}}-4x+1 \right)+2x\left( 3{{x}^{2}}-4x+1 \right)-3\left( 3{{x}^{2}}-4x+1 \right)$.
We will now simplify each of the terms on the right side of the above equation using the law of exponents which states that ${{x}^{a}}\times {{x}^{b}}={{x}^{a+b}}$.
Thus, we can simplify the expression $2{{x}^{3}}\left( 3{{x}^{2}}-4x+1 \right)$ and write is as $2{{x}^{3}}\left( 3{{x}^{2}}-4x+1 \right)=6{{x}^{5}}-8{{x}^{4}}+2{{x}^{3}}.....\left( 1 \right)$.
Similarly, we can simplify the expression $-3{{x}^{2}}\left( 3{{x}^{2}}-4x+1 \right)$ and write it as $-3{{x}^{2}}\left( 3{{x}^{2}}-4x+1 \right)=-9{{x}^{4}}+12{{x}^{3}}-3{{x}^{2}}.....\left( 2 \right)$.
Also, we can simplify the expression $2x\left( 3{{x}^{2}}-4x+1 \right)$ and write it as $2x\left( 3{{x}^{2}}-4x+1 \right)=6{{x}^{3}}-8{{x}^{2}}+2x.....\left( 3 \right)$.
Similarly, we can simplify the expression $-3\left( 3{{x}^{2}}-4x+1 \right)$ and write it as $-3\left( 3{{x}^{2}}-4x+1 \right)=-9{{x}^{2}}+12x-3.....\left( 4 \right)$.
Using equation (1), (2), (3), and (4), we have $\left( 2{{x}^{3}}-3{{x}^{2}}+2x-3 \right)\left( 3{{x}^{2}}-4x+1 \right)=6{{x}^{5}}-8{{x}^{4}}+2{{x}^{3}}-9{{x}^{4}}+12{{x}^{3}}-3{{x}^{2}}+6{{x}^{3}}-8{{x}^{2}}+2x-9{{x}^{2}}+12x-3$.
Further simplifying the above expression, we have $\left( 2{{x}^{3}}-3{{x}^{2}}+2x-3 \right)\left( 3{{x}^{2}}-4x+1 \right)=6{{x}^{5}}-17{{x}^{4}}+20{{x}^{3}}-20{{x}^{2}}+14x-3$.
Hence, the product of $\left( 2{{x}^{3}}-3{{x}^{2}}+2x-3 \right)$ and $\left( 3{{x}^{2}}-4x+1 \right)$ is $6{{x}^{5}}-17{{x}^{4}}+20{{x}^{3}}-20{{x}^{2}}+14x-3$, which is option (a).

Note: We observe that we multiply two polynomials (or algebraic expressions). We know the degree of the polynomial is the highest power of the exponent of the variable. One must keep in mind that the degree of product of two polynomials is the sum of the degree of both the polynomials. We observe that the degree of $\left( 2{{x}^{3}}-3{{x}^{2}}+2x-3 \right)$ is 3 and $\left( 3{{x}^{2}}-4x+1 \right)$ is 2. Thus, the degree of $6{{x}^{5}}-17{{x}^{4}}+20{{x}^{3}}-20{{x}^{2}}+14x-3$ is 5.