Math, asked by natashaghildiyal, 6 months ago

Simplify:
(2a2 + 5) (2a2 + 7) + 5​

Answers

Answered by vasukapoor001
1

Answer:

Given expression is: (a

3

−2a

2

+4a−5)−(−a

3

−8a+2a

2

+5)

=a

3

−2a

2

+4a−5+a

3

+8a−2a

2

−5

=2a

3

−4a

2

+12a−10

Hence simplified form of the given expression is =2a

3

−4a

2

+12a−10

Answered by Anonymous
1

Step-by-step explanation:

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Simplification or other simple results

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Enter an equation or a problem

We think you wrote:

2a2(5a-6)-5a(a2-3a+4)-7(a-5)

This deals with simplification or other simple results.

Overview

Steps

Topics

1 result(s) found

5a

3

+3a

2

−27a+35

See steps

Step by Step Solution:

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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).

STEP

1

:

Equation at the end of step 1

(((2•(a2))•(5a-6))-(5a•(((a2)-3a)+4)))-7•(a-5)

STEP

2

:

Trying to factor by splitting the middle term

2.1 Factoring a2-3a+4

The first term is, a2 its coefficient is 1 .

The middle term is, -3a its coefficient is -3 .

The last term, "the constant", is +4

Step-1 : Multiply the coefficient of the first term by the constant 1 • 4 = 4

Step-2 : Find two factors of 4 whose sum equals the coefficient of the middle term, which is -3 .

-4 + -1 = -5

-2 + -2 = -4

-1 + -4 = -5

1 + 4 = 5

2 + 2 = 4

4 + 1 = 5

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Equation at the end of step

2

:

(((2•(a2))•(5a-6))-5a•(a2-3a+4))-7•(a-5)

STEP

3

:

Equation at the end of step

3

:

((2a2•(5a-6))-5a•(a2-3a+4))-7•(a-5)

STEP

4

:

Equation at the end of step 4

(2a2•(5a-6)-5a•(a2-3a+4))-7•(a-5)

STEP

5

:

Checking for a perfect cube

5.1 5a3+3a2-27a+35 is not a perfect cube

Trying to factor by pulling out :

5.2 Factoring: 5a3+3a2-27a+35

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

Group 1: -27a+35

Group 2: 3a2+5a3

Pull out from each group separately :

Group 1: (-27a+35) • (1) = (27a-35) • (-1)

Group 2: (5a+3) • (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 :

5.3 Find roots (zeroes) of : F(a) = 5a3+3a2-27a+35

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 35.

The factor(s) are:

of the Leading Coefficient : 1,5

of the Trailing Constant : 1 ,5 ,7 ,35

Let us test ....

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

-1 1 -1.00 60.00

-1 5 -0.20 40.48

-5 1 -5.00 -380.00

-7 1 -7.00 -1344.00

-7 5 -1.40 64.96

-35 1 -35.00 -209720.00

1 1 1.00 16.00

1 5 0.20 29.76

5 1 5.00 600.00

7 1 7.00 1708.00

7 5 1.40 16.80

35 1 35.00 217140.00

Polynomial Roots Calculator found no rational roots

Final result :

5a3 + 3a2 - 27a + 35

Terms and topics

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Canceling out

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