Answer
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Hint: Here in this question, we are supposed to perform multiplication for polynomials. One should be aware about different types of polynomials. The rules of integers are also used for solving. The usage of algebraic identities will help to solve this question too.
Complete step-by-step solution:
When more than one term is in an expression it is called a polynomial. As we already know that there are various types of polynomials. First one is monomial, which contains only a single term. Second one is binomial, which contains two terms and third one is trinomial, which contains exactly three terms.
Let us know about the algebraic identities. Algebraic identities are just algebraic equations which stand true for every variable in them. There are various algebraic identities which exist. One of them is
\[\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right) = {x^3} - \left( {a + b + c} \right){x^2} + \left( {ab + bc + ca} \right)x - abc\] -----(1)
Now, let us solve the given question.
Given is$\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)$.
Using the algebraic identity shown in equation (1).
\[\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)\]and can be written as$\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)$. Hereby, we can say that\[a = 1,b = 2\]and\[c = 3\].
The terms on right-hand side of the identity in equation (1) are\[\left( {a + b + c} \right)\],\[\left( {ab + bc + ca} \right)\]and\[abc\]. Substituting the values of\[a,b\]and\[c\]in these terms and we get,
\[
\left( {a + b + c} \right) = \left( {1 + 2 + 3} \right) = 6 \\
\left( {ab + bc + ca} \right) = \left( {1 \times 2 + 2 \times 3 + 3 \times 1} \right) = \left( {2 + 6 + 3} \right) = 11 \\
\left( {abc} \right) = \left( {1 \times 2 \times 3} \right) = 6 \\
\]
Using the obtained values, we get,
\[\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right) = {x^3} - 6{x^2} + 11x - 6\]
Hence, the answer is\[{x^3} - 6{x^2} + 11x - 6\].
Note: Students are supposed to solve all the like terms until they get all the unlike terms. This will help in reducing the number of terms in the solution. It is always advisable to write terms in a decreasing order of their exponent. The rules of the integers should be taken care of while opening the brackets.
Complete step-by-step solution:
When more than one term is in an expression it is called a polynomial. As we already know that there are various types of polynomials. First one is monomial, which contains only a single term. Second one is binomial, which contains two terms and third one is trinomial, which contains exactly three terms.
Let us know about the algebraic identities. Algebraic identities are just algebraic equations which stand true for every variable in them. There are various algebraic identities which exist. One of them is
\[\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right) = {x^3} - \left( {a + b + c} \right){x^2} + \left( {ab + bc + ca} \right)x - abc\] -----(1)
Now, let us solve the given question.
Given is$\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)$.
Using the algebraic identity shown in equation (1).
\[\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)\]and can be written as$\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)$. Hereby, we can say that\[a = 1,b = 2\]and\[c = 3\].
The terms on right-hand side of the identity in equation (1) are\[\left( {a + b + c} \right)\],\[\left( {ab + bc + ca} \right)\]and\[abc\]. Substituting the values of\[a,b\]and\[c\]in these terms and we get,
\[
\left( {a + b + c} \right) = \left( {1 + 2 + 3} \right) = 6 \\
\left( {ab + bc + ca} \right) = \left( {1 \times 2 + 2 \times 3 + 3 \times 1} \right) = \left( {2 + 6 + 3} \right) = 11 \\
\left( {abc} \right) = \left( {1 \times 2 \times 3} \right) = 6 \\
\]
Using the obtained values, we get,
\[\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right) = {x^3} - 6{x^2} + 11x - 6\]
Hence, the answer is\[{x^3} - 6{x^2} + 11x - 6\].
Note: Students are supposed to solve all the like terms until they get all the unlike terms. This will help in reducing the number of terms in the solution. It is always advisable to write terms in a decreasing order of their exponent. The rules of the integers should be taken care of while opening the brackets.
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