Question

find if the second polynomial is a factor of the first polynomial using long division methods a= x'2-4x+8,x-2 b = 2x'3+x'2-3,2x+3 C=8x'4+10x'3-5x'2-4x+1,2x'2+x-1​

Answer 1

:

1

Simplify —

2

Equation at the end of step

1

:

1

((((((8•(x4))+(10•(x3)))-(5•(x2)))-4x)+(—•x2))+x)-1

2

STEP

2

:

Equation at the end of step 2

x2

((((((8•(x4))+(10•(x3)))-(5•(x2)))-4x)+——)+x)-1

2

STEP

3

:

Equation at the end of step

3

:

x2

((((((8•(x4))+(10•(x3)))-5x2)-4x)+——)+x)-1

2

STEP

4

:

Equation at the end of step

4

:

x2

((((((8•(x4))+(2•5x3))-5x2)-4x)+——)+x)-1

2

STEP

5

:

Equation at the end of step

5

:

x2

(((((23x4+(2•5x3))-5x2)-4x)+——)+x)-1

2

STEP

6

:

Rewriting the whole as an Equivalent Fraction

6.1 Adding a fraction to a whole

Rewrite the whole as a fraction using 2 as the denominator :

8x4 + 10x3 - 5x2 - 4x (8x4 + 10x3 - 5x2 - 4x) • 2

8x4 + 10x3 - 5x2 - 4x = ————————————————————— = ———————————————————————————

1 2

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

STEP

7

:

Pulling out like terms

7.1 Pull out like factors :

8x4 + 10x3 - 5x2 - 4x =

x • (8x3 + 10x2 - 5x - 4)

Checking for a perfect cube :

7.2 8x3 + 10x2 - 5x - 4 is not a perfect cube

Trying to factor by pulling out :

7.3 Factoring: 8x3 + 10x2 - 5x - 4

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

Group 1: -5x - 4

Group 2: 8x3 + 10x2

Pull out from each group separately :

Group 1: (5x + 4) • (-1)

Group 2: (4x + 5) • (2x2)

Bad news !! Factoring by pulling out fails :

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

Answer 2

Your first answer is that

Step-by-step explanation:

1= NO

2=YES

3=NO