`4q^3-6q^2-4q+3;2q-1`

Show that : (4p q+3q)^2-(4q p-3q)^2=48 p q^2 (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0

Let the polynomials be (1) -13q^(5) + 4q^(2) + 12q (2) (x^(2) + 4 ) ( x^(2) + 9) (3) 4q^(8) - q^(6) + q^(2) (4) - ( 5)/( 7) y^(12) + y^(3) + y^(5) Then ascending order of their degree is

Simplify ((q^(3-7)q xx q^(6-2q))/(q^(2q)xx q^(9-2q)))^((1)/(9))

If the difference of the roots of x^2-p x+q=0 is unity, then a) p^2+4q=1 b) p^2-4q=1 c) p^2+4q^2=(1+2q)^2 d) 4p^2+q^2=(1+2p)^2

Simplify ((q^(3-7q)xxq^(6-2q))/(q^(2q)xxq^(9-2q)))^(1/9) .

If 2p + 3q = 12 and 4p^(2) + 4pq - 3q^(2) = 126 , then what is the value of p + 2q ?

2p+3q=18 and 4p^(2)+4pq-3q^(2)-36=0 then what is (2p+q) equal to?

Resolve into factors : (4p-3q)^3+(3q-2)^3+(2-4p)^3