Math, asked by shraddha2444, 7 months ago

2+
D(x-3)(x+82)
a
b
IC
find a +7b
7b
5
7 b​

Answers

Answered by enriqueroll24
1

Answer:

STEP

1

:

Equation at the end of step 1

((b6) + 7b3) - 8 = 0

STEP

2

:

Trying to factor by splitting the middle term

2.1 Factoring b6+7b3-8

The first term is, b6 its coefficient is 1 .

The middle term is, +7b3 its coefficient is 7 .

The last term, "the constant", is -8

Step-1 : Multiply the coefficient of the first term by the constant 1 • -8 = -8

Step-2 : Find two factors of -8 whose sum equals the coefficient of the middle term, which is 7 .

-8 + 1 = -7

-4 + 2 = -2

-2 + 4 = 2

-1 + 8 = 7 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -1 and 8

b6 - 1b3 + 8b3 - 8

Step-4 : Add up the first 2 terms, pulling out like factors :

b3 • (b3-1)

Add up the last 2 terms, pulling out common factors :

8 • (b3-1)

Step-5 : Add up the four terms of step 4 :

(b3+8) • (b3-1)

Which is the desired factorization

Trying to factor as a Sum of Cubes:

2.2 Factoring: b3+8

Theory : A sum of two perfect cubes, a3 + b3 can be factored into :

(a+b) • (a2-ab+b2)

Proof : (a+b) • (a2-ab+b2) =

a3-a2b+ab2+ba2-b2a+b3 =

a3+(a2b-ba2)+(ab2-b2a)+b3=

a3+0+0+b3=

a3+b3

Check : 8 is the cube of 2

Check : b3 is the cube of b1

Factorization is :

(b + 2) • (b2 - 2b + 4)

Trying to factor by splitting the middle term

2.3 Factoring b2 - 2b + 4

The first term is, b2 its coefficient is 1 .

The middle term is, -2b its coefficient is -2 .

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 -2 .

-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

Trying to factor as a Difference of Cubes:

2.4 Factoring: b3-1

Theory : A difference of two perfect cubes, a3 - b3 can be factored into

(a-b) • (a2 +ab +b2)

Proof : (a-b)•(a2+ab+b2) =

a3+a2b+ab2-ba2-b2a-b3 =

a3+(a2b-ba2)+(ab2-b2a)-b3 =

a3+0+0+b3 =

a3+b3

Check : 1 is the cube of 1

Check : b3 is the cube of b1

Factorization is :

(b - 1) • (b2 + b + 1)

Trying to factor by splitting the middle term

2.5 Factoring b2 + b + 1

The first term is, b2 its coefficient is 1 .

The middle term is, +b its coefficient is 1 .

The last term, "the constant", is +1

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

Step-2 : Find two factors of 1 whose sum equals the coefficient of the middle term, which is 1 .

-1 + -1 = -2

1 + 1 = 2

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Equation at the end of step

2

:

(b+2)•(b2-2b+4)•(b-1)•(b2+b+1) = 0

STEP

3

:

Theory - Roots of a product

3.1 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.

Solving a Single Variable Equation:

3.2 Solve : b+2 = 0

Subtract 2 from both sides of the equation :

b = -2

Step-by-step explanation:

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