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Question: How do you factor \(p(x) = 2{x^3} - {x^2} - 18x + 9\) ?...

How do you factor p(x)=2x3−x2−18x+9p(x) = 2{x^3} - {x^2} - 18x + 9 ?

Explanation

Solution

Take out common terms from the first two terms and last two terms. Then write them together as the product of sums form. Now among the factorized terms, find the expression which can be further factorized and then write that also in the product of sums form. Now write all of them together and represent them as the factors of the given expression.

Formula used:
a2−b2=(a+b)(a−b){a^2} - {b^2} = (a + b)(a - b)

Complete step-by-step answer:
The given polynomial is p(x)=2x3−x2−18x+9p(x) = 2{x^3} - {x^2} - 18x + 9
The polynomial is of degree 33
Now, firstly let’s take the common terms out of the first two terms.
⇒p(x)=2x3−x2−18x+9\Rightarrow p(x) = 2{x^3} - {x^2} - 18x + 9
⇒p(x)=x2(2x−1)−18x+9\Rightarrow p(x) = {x^2}(2x - 1) - 18x + 9
Now secondly take the common terms out of the last two terms.
⇒p(x)=x2(2x−1)−9(2x−1)\Rightarrow p(x) = {x^2}(2x - 1) - 9(2x - 1)
Now upon writing it in the form of the product of sums, also known as factoring,
⇒p(x)=(x2−9)(2x−1)\Rightarrow p(x) = ({x^2} - 9)(2x - 1)
Now we can see that there’s still another polynomial left that can be factored further.
We can write x2−9{x^2} - 9 as x2−32{x^2} - {3^2}
And since, a2−b2=(a+b)(a−b){a^2} - {b^2} = (a + b)(a - b)
We can factorize it as,
⇒(x+3)(x−3)\Rightarrow (x + 3)(x - 3)
Now writing it all together we get,
⇒(x+3)(x−3)(2x−1)\Rightarrow (x + 3)(x - 3)(2x - 1)

∴p(x)=2x3−x2−18x+9\therefore p(x) = 2{x^3} - {x^2} - 18x + 9 can be factorized and the factors are, (x+3)(x−3)(2x−1)(x + 3)(x - 3)(2x - 1)

Additional information: The process of factorization is reverse multiplication. In the above question, we have multiplied two linear line equations to get a quadratic equation (a polynomial of degree 22 ) expression using the distributive law. The name quadratic comes from “quad” meaning square, because the variable gets squared. Also known as a polynomial of degree 22 .

Note:
Whenever there is a polynomial that is to be solved, the solution contains the roots of the expression. The number of roots is decided by the degree of the polynomial. You can always cross-check your answer by placing the value of xx (factors of the expression) back in the equation. If you get LHS = RHS then your answer is correct.