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
Answer:
1. -5 and -8
2. -7 and -4
3. -15 and -1
4. 7 and -2
5. -5 and 1
6. 2 and -5
7. -8 and 9
8. 5 and -4
Step-by-step explanation:
Answer:
you save $5, i can't send a diagram to help though.
Step-by-step explanation:
Answer:
$14.30
Step-by-step explanation:
26 - 11.70 = 14.30
Answer:
1 false
2 true
3 true
4 false
5 true
Step-by-step explanation:
f(a) = (2a - 7 + a^2) and g(a) = (5 – a).
1 false f(a) is a second degree polynomial and g(a) is a first degree polynomial
When added together, they will be a second degree polynomial
2. true When we add and subtract polynomials, we still get a polynomial, so it is closed under addition and subtraction
3. true f(a) + g(a) = (2a - 7 + a^2) + (5 – a)
Combining like terms = a^2 +a -2
4. false f(a) - g(a) = (2a - 7 + a^2) - (5 – a)
Distributing the minus sign (2a - 7 + a^2) - 5 + a
Combining like terms a^2 +3a -12
5. true f(a)* g(a) = (2a - 7 + a^2) (5 – a).
Distribute
(2a - 7 + a^2) (5) – (2a - 7 + a^2) (a)
10a -35a +5a^2 -2a^2 -7a +a^3
Combining like term
-a^3 + 3 a^2 + 17 a - 35
