Math, asked by sudiptodas123d2h, 7 months ago


5a3 + 11a2 + 4a - 2​

Answers

Answered by Prakharbro
1

Step-by-step explanation:

Changes made to your input should not affect the solution:

(1): "a2" was replaced by "a^2". 1 more similar replacement(s).

STEP

1

:

Equation at the end of step 1

(((5 • (a3)) + 11a2) + 4a) - 2

STEP

2

:

Equation at the end of step

2

:

((5a3 + 11a2) + 4a) - 2

STEP

3

:

Checking for a perfect cube

3.1 5a3+11a2+4a-2 is not a perfect cube

Trying to factor by pulling out :

3.2 Factoring: 5a3+11a2+4a-2

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1: 4a-2

Group 2: 5a3+11a2

Pull out from each group separately :

Group 1: (2a-1) • (2)

Group 2: (5a+11) • (a2)

Bad news !! Factoring by pulling out fails :

The groups have no common factor and can not be added up to form a multiplication.

Polynomial Roots Calculator :

3.3 Find roots (zeroes) of : F(a) = 5a3+11a2+4a-2

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 5 and the Trailing Constant is -2.

The factor(s) are:

of the Leading Coefficient : 1,5

of the Trailing Constant : 1 ,2

Let us test ....

P Q P/Q F(P/Q) Divisor

-1 1 -1.00 0.00 a+1

-1 5 -0.20 -2.40

-2 1 -2.00 -6.00

-2 5 -0.40 -2.16

1 1 1.00 18.00

1 5 0.20 -0.72

2 1 2.00 90.00

2 5 0.40 1.68

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

5a3+11a2+4a-2

can be divided with a+1

Polynomial Long Division :

3.4 Polynomial Long Division

Dividing : 5a3+11a2+4a-2

("Dividend")

By : a+1 ("Divisor")

dividend 5a3 + 11a2 + 4a - 2

- divisor * 5a2 5a3 + 5a2

remainder 6a2 + 4a - 2

- divisor * 6a1 6a2 + 6a

remainder - 2a - 2

- divisor * -2a0 - 2a - 2

remainder 0

Quotient : 5a2+6a-2 Remainder: 0

Trying to factor by splitting the middle term

3.5 Factoring 5a2+6a-2

The first term is, 5a2 its coefficient is 5 .

The middle term is, +6a its coefficient is 6 .

The last term, "the constant", is -2

Step-1 : Multiply the coefficient of the first term by the constant 5 • -2 = -10

Step-2 : Find two factors of -10 whose sum equals the coefficient of the middle term, which is 6 .

-10 + 1 = -9

-5 + 2 = -3

-2 + 5 = 3

-1 + 10 = 9

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Final result :

(5a2 + 6a - 2) • (a + 1)

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