Show that [tex]P(2)=0[/tex] for [tex]P(x)=2 x^3-x^2-6 x[/tex].
Answer (2)
Substitute x = 2 into P ( x ) = 2 x 3 − x 2 − 6 x and calculate P ( 2 ) = 2 ( 2 ) 3 − ( 2 ) 2 − 6 ( 2 ) = 16 − 4 − 12 = 0 .
Factor out x from P ( x ) to get P ( x ) = x ( 2 x 2 − x − 6 ) .
Factor the quadratic 2 x 2 − x − 6 as ( 2 x + 3 ) ( x − 2 ) .
The simplified form of the polynomial is P ( x ) = x ( 2 x + 3 ) ( x − 2 ) .
P ( 2 ) = 0 , P ( x ) = x ( 2 x + 3 ) ( x − 2 )
Explanation
Understanding the Problem We are given the polynomial P ( x ) = 2 x 3 − x 2 − 6 x and we want to show that P ( 2 ) = 0 . This means we need to substitute x = 2 into the polynomial and verify that the result is indeed 0.
Substituting and Calculating Now, let's substitute x = 2 into the polynomial: P ( 2 ) = 2 ( 2 ) 3 − ( 2 ) 2 − 6 ( 2 ) P ( 2 ) = 2 ( 8 ) − 4 − 12 P ( 2 ) = 16 − 4 − 12 P ( 2 ) = 12 − 12 P ( 2 ) = 0 Thus, we have shown that P ( 2 ) = 0 .
Simplifying the Polynomial Next, we need to simplify the polynomial P ( x ) = 2 x 3 − x 2 − 6 x . We can do this by factoring out a common factor of x from each term: P ( x ) = x ( 2 x 2 − x − 6 ) Now, we need to factor the quadratic expression 2 x 2 − x − 6 . We are looking for two numbers that multiply to 2 × − 6 = − 12 and add up to − 1 . These numbers are − 4 and 3 . So we can rewrite the quadratic as: 2 x 2 − x − 6 = 2 x 2 − 4 x + 3 x − 6 Now, we can factor by grouping: 2 x 2 − 4 x + 3 x − 6 = 2 x ( x − 2 ) + 3 ( x − 2 ) = ( 2 x + 3 ) ( x − 2 ) Therefore, the factored form of the polynomial is: P ( x ) = x ( 2 x + 3 ) ( x − 2 )
Examples
Polynomials are used to model various real-world phenomena, such as the trajectory of a projectile, the growth of a population, or the behavior of electrical circuits. Verifying roots of polynomials, like we did with P(2) = 0, helps us find specific values where these models have particular behaviors, such as when a projectile hits the ground or when a circuit reaches a certain state. Factoring polynomials simplifies these models, making them easier to analyze and understand.
We verified that P ( 2 ) = 0 by substituting x = 2 into P ( x ) = 2 x 3 − x 2 − 6 x , which resulted in zero. Additionally, we factored the polynomial as P ( x ) = x ( 2 x + 3 ) ( x − 2 ) , confirming that x = 2 is indeed a root. Thus, the polynomial has a root at 2 .
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