Polynomial Multiplication We learnt about polynomials
here . Today we will see how two or more polynomials can be multiplied.
We know that multiplication is distributive over addition/subtraction for numbers. In case of polynomials, each term of a polynomial, as well as the polynomial itself, is actually a number. Based on that when we multiply two polynomials, we can distribute any one of them over all the term.
If we have polynomials:
$P(x) = a_nx^{n} + a_{n-1}x^{n-1} + a_{n-2}x^{n-2}+\ldots \ + a_0$, and
$G(x) = c_nx^{n} + c_{n-1}x^{n-1} + c_{n-2}x^{n-2}+\ldots \ + c_0$
then we can multiply the two polynomials using the following steps:
$P(x) \times G(x)$
$= \left(a_nx^{n} + a_{n-1}x^{n-1} + a_{n-2}x^{n-2}+\ldots \ + a_0 \right) \times g(x) \phantom{000} \Leftarrow \hbox{replacing the value of p(x)}$
$= a_nx^{n}\times g(x) + a_{n-1}x^{n-1}\times g(x) + a_{n-2}x^{n-2}\times g(x)+\ldots \ + a_0\times g(x) \phantom{000} \Leftarrow \hbox{g(x) got distributed over all terms of p(x)}$
$= a_nx^{n}\times \left( c_nx^{n} + c_{n-1}x^{n-1} + c_{n-2}x^{n-2}+\ldots \ + c_0 \right) + a_{n-1}x^{n-1}\times \left( c_nx^{n} + c_{n-1}x^{n-1} + c_{n-2}x^{n-2}+\ldots \ + c_0 \right)$$\phantom{000} + a_{n-2}x^{n-2}\times \left( c_nx^{n} + c_{n-1}x^{n-1} + c_{n-2}x^{n-2}+\ldots \ + c_0 \right)+\ldots \ $$\phantom{000}+ a_0\times \left( c_nx^{n} + c_{n-1}x^{n-1} + c_{n-2}x^{n-2}+\ldots \ + c_0 \right) \phantom{000} \Leftarrow \hbox{replaced g(x) with its terms}$
$= a_nx^{n}\times c_nx^{n} + a_nx^{n}\times c_{n-1}x^{n-1}+\ldots \ + a_0 \times c_0\phantom{000} \Leftarrow \hbox{we opened all the brackets and distributed the outside terms}$
Now, we are ready for an example of this.
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Find the value of $p(x) \times g(x)$ where,
$P(x) = 8x^3 - 12x^2 + 4x + 7$, and
$G(x) = 3x^2 + 9x - 5$
$P(x) \times G(x)$
$= \left(8x^3 - 12x^2 + 4x + 7 \right) \times \left(3x^2 + 9x - 5 \right)$
$= 8x^3 \left(3x^2 + 9x - 5 \right) - 12x^2\left(3x^2 + 9x - 5 \right) + 4x\left(3x^2 + 9x - 5 \right)$ $\ \ \ \ + 7\left(3x^2 + 9x - 5 \right)$
$= 24x^5 + 72x^4 - 40x^3 - 36x^4 - 108x^3 + 60x^2 + 12x^3 + 36x^2$ $\ \ \ \ \ - 20x + 21x^2 + 63x - 35$
$= 24x^5 + 72x^4 - 36x^4 - 40x^3 - 108x^3 + 12x^3 + 60x^2 + 36x^2 + 21x^2$ $\ \ \ \ \ - 20x + 63x - 35$
$= 24x^5 + 36x^4 - 136x^3 + 117x^2 + 43x - 35$
Now let us try the following problem:
--------- Reference to question: da07567c-6e03-48d4-8190-09c8398201fd ---------