If x=y and y=10, then x=10 exemplifies which property?
1 answer:
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The factorization of given cubic polynomial 6x³ - 11x² - 12x + 5 is:
6x³ - 11x² - 12x + 5 = (x + 1)(x - )(x -
)
For given question,
We have been given the cubic polynomial 6x³ - 11x² - 12x + 5
We need to factorize given cubic polynomial.
By the rational roots theorem, any rational zero of f(x) is expressible in the form ± for integers p, q with p a divisor of the constant term 5 and q a divisor of the coefficient 6 of the leading term.
Factors of p = 5: 1, 5
Factors of q = 6: 1, 2, 3, 6
That means that the only possible rational zeros are:
±{ }
= ±{ }
We need to find the exact zeros of given cubic polynomial.
For x = 1,
6(1)³ - 11(1)² - 12(1) + 5 = -12
This means, x = 1 is not a zero of given cubic polynomial.
For x = -1,
6(-1)³ - 11(-1)² - 12(-1) + 5 = 0
This means, x = -1 is a zero of given cubic polynomial and (x + 1) is a factor.
To factorize given cubic polynomial we use synthetic division.
The synthetic division (6x³ - 11x² - 12x + 5) ÷ (x + 1) is as shown in following image.
⇒ 6x³ - 11x² - 12x + 5 = (x + 1)(6x² - 17x + 5)
The factors of above quadratic polynomial 6x² - 17x + 5 are:
⇒ 6x² - 17x + 5 = (x - )(x -
)
So, the factors of given cubic polynomial are:
⇒ 6x³ - 11x² - 12x + 5 = (x + 1)(x - )(x -
)
Therefore, the factorization of given cubic polynomial 6x³ - 11x² - 12x + 5 is: 6x³ - 11x² - 12x + 5 = (x + 1)(x - )(x -
)
Learn more about the polynomial here:
#SPJ4
Answer:
According to steps 2 and 4. The second-order polynomial must be added by and
to create a perfect square trinomial.
Step-by-step explanation:
Let consider a second-order polynomial of the form ,
. The procedure is presented below:
1) (Given)
2) (Compatibility with addition/Existence of additive inverse/Modulative property)
3) (Compatibility with multiplication)
4) (Compatibility with addition/Existence of additive inverse/Modulative property)
5) (Perfect square trinomial)
According to steps 2 and 4. The second-order polynomial must be added by and
to create a perfect square trinomial.