Find a if a³+6a²+12a =504
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Step-by-step explanation:
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "a2" was replaced by "a^2". 1 more similar replacement(s).
STEP1:Equation at the end of
(((a3) + (2•3a2)) + 12a) + 7
STEP2:Checking for a perfect cube 2.1 a3+6a2+12a+7 is not a perfect cube
Trying to factor by pulling out :
2.2 Factoring: a3+6a2+12a+7
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: a3+7
Group 2: 6a2+12a
Pull out from each group separately :
Group 1: (a3+7) • (1)
Group 2: (a+2) • (6a)
The groups have no common factor and can not be added up to form a multiplication.
Polynomial Roots Calculator :
2.3 Find roots (zeroes) of : F(a) = a3+6a2+12a+7
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 7.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,7
Let us test ....
P Q P/Q F(P/Q) Divisor
-1 1 -1.00 0.00 a+1
-7 1 -7.00 -126.00
1 1 1.00 26.00
7 1 7.00 728.00
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+6a2+12a+7
can be divided with a+1
Polynomial Long Division :
2.4 Polynomial Long Division
Dividing : a3+6a2+12a+7
("Dividend")
By : a+1 ("Divisor")
dividend a3 + 6a2 + 12a + 7
- divisor * a2 a3 + a2
remainder 5a2 + 12a + 7
- divisor * 5a1 5a2 + 5a
remainder 7a + 7
- divisor * 7a0 7a + 7
remainder 0
Quotient : a2+5a+7 Remainder: 0
Trying to factor by splitting the middle term
2.5 Factoring a2+5a+7
The first term is, a2 its coefficient is 1 .
The middle term is, +5a its coefficient is 5 .
The last term, "the constant", is +7
Step-1 : Multiply the coefficient of the first term by the constant 1 • 7 = 7
Step-2 : Find two factors of 7 whose sum equals the coefficient of the middle term, which is 5 .
-7 + -1 = -8
-1 + -7 = -8
1 + 7 = 8
7 + 1 = 8
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Final result :
(a2 + 5a + 7) • (a + 1)
Hope its help...
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