factories a3-2a-4 =0
Answers
Answer:
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Find roots (zeroes) of : F(a) = a3-2a-4
Polynomial Roots Calculator is a set of methods aimed at finding values of a for which F(a)=0
Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers a which can be expressed as the quotient of two integers
The Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient
In this case, the Leading Coefficient is 1 and the Trailing Constant is -4.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,4
Let us test .... P Q P/Q F(P/Q) Divisor -1 1 -1.00 -3.00 -2 1 -2.00 -8.00 -4 1 -4.00 -60.00 1 1 1.00 -5.00 2 1 2.00 0.00 a-2 4 1 4.00 52.00
The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms
In our case this means that
a3-2a-4
can be divided with a-2
Polynomial Long Division :
1.2 Polynomial Long Division
Dividing : a3-2a-4
("Dividend")
By : a-2 ("Divisor")
dividend a3 - 2a - 4 - divisor * a2 a3 - 2a2 remainder 2a2 - 2a - 4 - divisor * 2a1 2a2 - 4a remainder 2a - 4 - divisor * 2a0 2a - 4 remainder 0
Quotient : a2+2a+2 Remainder: 0
Trying to factor by splitting the middle term
1.3 Factoring a2+2a+2
The first term is, a2 its coefficient is 1 .
The middle term is, +2a its coefficient is 2 .
The last term, "the constant", is +2
Step-1 : Multiply the coefficient of the first term by the constant 1 • 2 = 2
Step-2 : Find two factors of 2 whose sum equals the coefficient of the middle term, which is 2 .
-2 + -1 = -3 -1 + -2 = -3 1 + 2 = 3 2 + 1 = 3
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Equation at the end of step1:
(a2 + 2a + 2) • (a - 2) = 0
STEP2:Theory - Roots of a product
2.1 A product of several terms equals zero.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term = 0 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.