9p4r4-6p4q2r2+p8 factorize
Answers
Answer:
9q4r4-6p4q2r2+p8
Final result :
(3q2r2 - p4)2
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "p8" was replaced by "p^8". 5 more similar replacement(s).
Step by step solution :
Step 1 :
Equation at the end of step 1 :
(((9•(q4))•(r4))-(((2•3p4)•q2)•r2))+p8
Step 2 :
Equation at the end of step 2 :
((32q4 • r4) - (2•3q2r2p4)) + p8
Step 3 :
Trying to factor a multi variable polynomial :
3.1 Factoring 9q4r4 - 6q2r2p4 + p8
Try to factor this multi-variable trinomial using trial and error
Found a factorization : (3q2r2 - p4)•(3q2r2 - p4)
Detecting a perfect square :
3.2 9q4r4 -6q2r2p4 +p8 is a perfect square
It factors into (3q2r2-p4)•(3q2r2-p4)
which is another way of writing (3q2r2-p4)2
How to recognize a perfect square trinomial:
• It has three terms
• Two of its terms are perfect squares themselves
• The remaining term is twice the product of the square roots of the other two terms
Trying to factor as a Difference of Squares :
3.3 Factoring: 3q2r2-p4
Put the exponent aside, try to factor 3q2r2-p4
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 : 3 is not a square !!
Ruling : Binomial can not be factored as the
difference of two perfect squares
Final result :
(3q2r2 - p4)2