Given a polynomial 
and a point 
, we have that
We know that our cubic function is zero at -4, 0 and 5, which means that our polynomial is a multiple of
Since this is already a cubic polynomial (it's the product of 3 polynomials with degree one), we can only adjust a multiplicative factor: our function must be
To fix the correct value for a, we impose 
:
And so we must impose
So, the function we're looking for is
Yep it is greater. Hope this helps
Answer:
p²q³ + pq and pq(pq² + 1)
Step-by-step explanation:
Given
3p²q² - 3p²q³ +4p²q³ -3p²q² + pq
Required
Collect like terms
We start by rewriting the expression
3p²q² - 3p²q³ +4p²q³ -3p²q² + pq
Collect like terms
3p²q² -3p²q² - 3p²q³ +4p²q³ + pq
Group like terms
(3p²q² -3p²q²) - (3p²q³ - 4p²q³ ) + pq
Perform arithmetic operations on like terms
(0) - (-p²q³) + pq
- (-p²q³) + pq
Open bracket
p²q³ + pq
The answer can be further simplified
Factorize p²q³ + pq
pq(pq² + 1)
Hence, 3p²q² - 3p²q³ +4p²q³ -3p²q² + pq is equivalent to p²q³ + pq and pq(pq² + 1)
Standard form is y=mx+b
y=9x+4
Consecutive= One after another
17 + 18 = 35
Your two consecutive numbers are 17 and 18
