8. Find the square root of x^6+ 8x^4 - 2x^3 + 16x^2– 8x + 1 using division method. answer pls now
Answers
Answer: ( ^ 3 + 4 − 1 ) ^2
and alsoo see the picture attached
Step-by-step explanation:
yeah sure the answer is is is is ( ^ 3 + 4 − 1 ) ^2
x^6+8x^4-2x^3+16x^2–8x+1
here below how to solve your problem
step 1: find one factor
factor by grouping
(x^3+4x-1)(x^3+4x-1)
step 2: find one factor
( ^ 3 + 4 − 1 ) ^2
so the answer is ( ^ 3 + 4 − 1 ) ^2
another solution below hope it helps you
Step by Step Solution
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Step by step solution :
STEP
1
:
Equation at the end of step 1
((x6) - (8 • (x4))) - 24x2 = 0
STEP
2
:
Equation at the end of step
2
:
((x6) - 23x4) - 24x2 = 0
STEP
3
:
STEP
4
:
Pulling out like terms
4.1 Pull out like factors :
x6 - 8x4 - 16x2 = x2 • (x4 - 8x2 - 16)
Trying to factor by splitting the middle term
4.2 Factoring x4 - 8x2 - 16
The first term is, x4 its coefficient is 1 .
The middle term is, -8x2 its coefficient is -8 .
The last term, "the constant", is -16
Step-1 : Multiply the coefficient of the first term by the constant 1 • -16 = -16
Step-2 : Find two factors of -16 whose sum equals the coefficient of the middle term, which is -8 .
-16 + 1 = -15
-8 + 2 = -6
-4 + 4 = 0
-2 + 8 = 6
-1 + 16 = 15
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Equation at the end of step
4
:
x2 • (x4 - 8x2 - 16) = 0
STEP
5
:
Theory - Roots of a product
5.1 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:
5.2 Solve : x2 = 0
Solution is x2 = 0
Solving a Single Variable Equation:
Equations which are reducible to quadratic :
5.3 Solve x4-8x2-16 = 0
This equation is reducible to quadratic. What this means is that using a new variable, we can rewrite this equation as a quadratic equation Using w , such that w = x2 transforms the equation into :
w2-8w-16 = 0
Solving this new equation using the quadratic formula we get two real solutions :
9.6569 or -1.6569
Now that we know the value(s) of w , we can calculate x since x is √ w
Doing just this we discover that the solutions of
x4-8x2-16 = 0
are either :
x =√ 9.657 = 3.10755 or :
x =√ 9.657 = -3.10755 or :
x =√-1.657 = 0.0 + 1.28719 i or :
x =√-1.657 = 0.0 - 1.28719 i
5 solutions were found :
x =√-1.657 = 0.0 - 1.28719 i
x =√-1.657 = 0.0 + 1.28719 i
x =√ 9.657 = -3.10755
x =√ 9.657 = 3.10755
x2 = 0
