<span>3x^2y^2 − 2xy^2 − 8y^2 =
y^2 (3x^2 - 2x - 8) =
factoring with leading coefficient:
for ax2+bx+c find two numbers n,m, that m*n = a*c and m+n = b
</span><span><span>
3x^2 - 2x - 8
a=3, b=-2, c=-8
</span>a*c = 3*(-8) = -24
-24=(-6)*4 and -6+4=-2, so m=-6 and n=4
replace bx with mx + nx and factor by grouping
</span><span>
3x^2 - 2x - 8 = </span>3x^2 -6x + 4x -8 = 3x(x-2) + 4(x-2) = (3x+4)(x-2)
answer:
<span>3x^2y^2 − 2xy^2 − 8y^2 = y^2(3x+4)(x-2)</span>
The rational root theorem states that the rational roots of a polynomial can only be in the form p/q, where p divides the constant term, and q divides the leading term.
In your case, both the leading term 5 and the constant term 11 are primes, so their only divisors are 1 and themselves.
So, the only feasible solutions are
For the record, in this case, none of the feasible solutions are actually a root of the polynomial.
<span>the blank boxes are for you to plug in x=20 to prove its right.
so it would be
3(20)-4=2(20+8)
60-4=2(28)
56=56
so its true!</span>
Answer:
id say a
Step-by-step explanation:
Answer:c
Step-by-step explanation:
The Answer is C.
