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Mandarinka [93]
3 years ago
9

Find the answer to the question please 16t^4/8t

Mathematics
1 answer:
sammy [17]3 years ago
4 0

Answer:

STEP

1

:

Equation at the end of step 1

 ((16 • (t4)) -  23t2) +  1

STEP

2

:

Equation at the end of step

2

:

 (24t4 -  23t2) +  1

STEP

3

:

Trying to factor by splitting the middle term

3.1     Factoring  16t4-8t2+1

The first term is,  16t4  its coefficient is  16 .

The middle term is,  -8t2  its coefficient is  -8 .

The last term, "the constant", is  +1

Step-1 : Multiply the coefficient of the first term by the constant   16 • 1 = 16

Step-2 : Find two factors of  16  whose sum equals the coefficient of the middle term, which is   -8 .

     -16    +    -1    =    -17

     -8    +    -2    =    -10

     -4    +    -4    =    -8    That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -4  and  -4

                    16t4 - 4t2 - 4t2 - 1

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

                   4t2 • (4t2-1)

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

                    1 • (4t2-1)

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

                   (4t2-1)  •  (4t2-1)

            Which is the desired factorization

Trying to factor as a Difference of Squares:

3.2      Factoring:  4t2-1

Theory : A difference of two perfect squares,  A2 - B2  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =

        A2 - AB + BA - B2 =

        A2 - AB + AB - B2 =

        A2 - B2

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check :  4  is the square of  2

Check : 1 is the square of 1

Check :  t2  is the square of  t1

Factorization is :       (2t + 1)  •  (2t - 1)

Trying to factor as a Difference of Squares:

3.3      Factoring:  4t2 - 1

Check :  4  is the square of  2

Check : 1 is the square of 1

Check :  t2  is the square of  t1

Factorization is :       (2t + 1)  •  (2t - 1)

Multiplying Exponential Expressions:

3.4    Multiply  (2t + 1)  by  (2t + 1)

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is  (2t+1)  and the exponents are :

         1 , as  (2t+1)  is the same number as  (2t+1)1

and   1 , as  (2t+1)  is the same number as  (2t+1)1

The product is therefore,  (2t+1)(1+1) = (2t+1)2

Multiplying Exponential Expressions:

3.5    Multiply  (2t-1)  by  (2t-1)

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is  (2t-1)  and the exponents are :

         1 , as  (2t-1)  is the same number as  (2t-1)1

and   1 , as  (2t-1)  is the same number as  (2t-1)1

The product is therefore,  (2t-1)(1+1) = (2t-1)2

Final result :

 (2t + 1)2 • (2t - 1)2

Step-by-step explanation:

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