p^2- 1/2 pq + 3p^2 - 3/2 pq - 3p^2 can anyone answer this plss:)
Answers
Answer:
Simplify —
p
Equation at the end of step
1
:
q q
(((((3•(p2))+(3p•————))-(q2))-3p)-(2•—))-q
(p2) p
STEP
2
:
q
Simplify ——
p2
Equation at the end of step
2
:
q 2q
(((((3•(p2))+(3p•——))-q2)-3p)-——)-q
p2 p
STEP
3
:
Equation at the end of step
3
:
3q 2q
((((3p2 + ——) - q2) - 3p) - ——) - q
p p
STEP
4
:
Rewriting the whole as an Equivalent Fraction
4.1 Adding a fraction to a whole
Rewrite the whole as a fraction using p as the denominator :
3p2 3p2 • p
3p2 = ——— = ———————
1 p
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
Adding fractions that have a common denominator :
4.2 Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:
3p2 • p + 3q 3p3 + 3q
———————————— = ————————
p p
Equation at the end of step
4
:
(3p3 + 3q) 2q
(((—————————— - q2) - 3p) - ——) - q
p p
STEP
5
:
Rewriting the whole as an Equivalent Fraction :
5.1 Subtracting a whole from a fraction
Rewrite the whole as a fraction using p as the denominator :
q2 q2 • p
q2 = —— = ——————
1 p
STEP
6
:
Pulling out like terms :
6.1 Pull out like factors :
3p3 + 3q = 3 • (p3 + q)
Trying to factor as a Sum of Cubes:
6.2 Factoring: p3 + q
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 : p3 is the cube of p1
Check : q 1 is not a cube !!
Ruling : Binomial can not be factored as the difference of two perfect cubes
Adding fractions that have a common denominator :
6.3 Adding up the two equivalent fractions
3 • (p3+q) - (q2 • p) 3p3 - pq2 + 3q
————————————————————— = ——————————————
p p
Equation at the end of step
6
:
(3p3 - pq2 + 3q) 2q
((———————————————— - 3p) - ——) - q
p p
STEP
7
:
Rewriting the whole as an Equivalent Fraction :
7.1 Subtracting a whole from a fraction
Rewrite the whole as a fraction using p as the denominator :
3p 3p • p
3p = —— = ——————
1 p
Trying to factor a multi variable polynomial :
7.2 Factoring 3p3 - pq2 + 3q
Try to factor this multi-variable trinomial using trial and error
Factorization fails
Adding fractions that have a common denominator :
7.3 Adding up the two equivalent fractions
(3p3-pq2+3q) - (3p • p) 3p3 - 3p2 - pq2 + 3q
——————————————————————— = ————————————————————
p p
Equation at the end of step
7
:
(3p3 - 3p2 - pq2 + 3q) 2q
(—————————————————————— - ——) - q
p p
STEP
8
:
Checking for a perfect cube :
8.1 3p3-3p2-pq2+3q is not a perfect cube
Adding fractions which have a common denominator :
8.2 Adding fractions which have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:
(3p3-3p2-pq2+3q) - (2q) 3p3 - 3p2 - pq2 + q
——————————————————————— = ———————————————————
p p
Equation at the end of step
8
:
(3p3 - 3p2 - pq2 + q)
————————————————————— - q
Answer:
hi mate
here is it
good afternoon
