Let's calculate the values of $p(7)$ and $p(3)$ :

$p(7)=7^2a+7b+c = 49a + 7b+c$

$p(3)=3^2a+3b+c=9a+3b+c$

Subtracting, we have $p(7) - p(3) = 49a+7b+c-9a-3b-c = 40a + 4 b$

Try subtracting $p(9) - p(1)$ too: $p(9) - p(1) = 81a+9b+c-a-b-c = 80a + 8 b$

This is getting us nowhere, right? Look again at the two values. Notice that $2(40a+4b)=80a+8b$ , so the value of $p(9)-p(1)$ is double the value of $p(7) - p(3)$ .

Substititing, we have $p(9) - p(1) = 2 \cdot 214 = \boxed{428}$

I ignored this entirely:

$p(x) = ax^{2} + bx + c$

$p(7) - p(3) = 214$ $p(7) - p(3) = p(4)$ Therefore...

$p(4) = 214$

$p(9) - p(1) = p(8)$

$p(4) \times 2 = p(8)$

Therefore...

$p(8) = 214 \times 2 = 428$

So the answer is 428 .

From the question, we can say

$p(7) - p(3)$ $=$ $214$

$7^{2}a + 7b + c - (3^{2}a + 3b +c)$ $=$ $214$

$49a + 7b + c - (9a + 3b +c)$ $=$ $214$

$40a + 4b$ $=$ $214$

Then,

$p(9) - p(1)$

$=$ $9^{2}a + 9b + c - (1^{2}a + 1b + c)$

$=$ $81a + 9b + c - (a + b + c)$

$=$ $80a + 8b$

$=$ $2(40a + 4b)$

$=$ $2(214)$

$=$ $\boxed {428}$

1) We have $p(x) = ax^2 + bx + c$ and $p(7) - p(3) = 214$ .

2) The value of $p(7)$ , when $x = 7$ in $p(x) = ax^2 + bx + c$ :

$p(7) = 7^{2}a + 7b + c$

$\implies 49a + 7b + c$

3) And the value of $p(3)$ , when $x= 3$ in $p(x) = ax^2 + bx + c$ :

$p(3) = 3^{2}a + 3b + c$

$\implies 9a + 3b + c$

4) Work out the value of $p(7) - p(3)$ :

$p(7) - p(3) = (49a + 7b + c) - (9a + 3b + c)$

$\implies 40a + 4b$

$\implies 4(10a + b)$

5) Putting the value of $p(7) - p(3) = 214$ :

$p(7) - p(3) = 214$

$\implies 4(10a + b) = 214$

$\implies 10a + b = \frac{214}{4}$

6) We get the value of $10a + b = \frac{214}{4}$

7) $p(9) - p(1)$ should be [like step 2 or 3] :

$p(9) - p(1) = (9^{2}a + 9b + c) - (1^{2}a + 1.b + c)$

$\implies (81a + 9b + c) - (a + b + c)$

$\implies 80a + 8b$

$\implies 8(10a + b)$

8) Putting the value of $10a + b = \frac{214}{4}$ [from step 6] :

$p(9) - p(1) = 8(10a + b)$

$\implies 8 \times \frac{214}{4}$

$\implies 2 \times 214 \implies 428$

9) We get the value of $p(9) - p(1) = \boxed{428}$

p(7) - p(3) = 40a + 4b

p(9) - p(1) = 80a + 8b

Since 80a + 8b is double 40a + 4b you just need to double 214 to get 428

p(7) = 49a + 7b + c and p(3) = 9a + 3b + c therefore, p(7) - p(3) = 40a + 4b = 214, and then p(9) = 81a + 9b + c and p(1) = a+b+c and p(9) - p(1) = 80a + 8b = 2(40a+4b) = 2(214) = 428

If you plug 7 and 3 into the equation you get:

[a(49) - b(7) + c] - [a(9) + b(3) + c] = a(40) - b(4)

> If you do the same for P(9) - P(1)

You get: a(80) + b(8)

= 2 [ a(40) + b(4) ] = 2 * 214 = 428

By the definition we have that $p(7)=49a+7b+c,$ and $p(3)=9a+3b+c,$ so $p(7)-p(3)=40a+4b=4(10a+b)=214.$ On the other hand we have $p(9)=81a+9b+c,$ and $p(1)=a+b+c,$ so $p(9)-p(1)=80a+8b=8(10a+b)=2(4(10a+b)).$ But $4(10a+b)=214.$ Hence $p(9)-p(1)=2(214)=\boxed{428}.$

First off I looked for the constant value, the first thing that struck out was that it was the same function both times. I noted this and then inserted the variables into the equation giving; P(7)=49a+7b+c P(3)=9a+3b+c I then inserted these equations into the second equation giving; (49a+7b+c)-(9a+3b+c)=214 Solving gives; 40a+4b=214 I then did the same for the second set of values giving; 80a+8b=x Since it is the same function and we have a linear equation they are directly proportional, e.g if one doubles the other doubles which is what is happening here so; 1/2x=214 x=428

On solving p(7) - p(3), we arrive at the equation, 40a + 4b = 214. Now, to get to know more about the constants, we need at least two more equations. The present situation seems to be against this idea. So lets keep the deduced equation aside and try to solve p(9) - p(1). Now, on solving this, we end up at 80a + 8b, which is interestingly the double of (40a + 4b). Since we have already found out this value, we multiply it(i.e. 214 ) by 2 to get 428.

We have, p(7)-p(3)=(49a+7b+c)-(9a+3b+c)=40a+4b=214 (given)............................Then, p(9)-p(1)=(81a+9b+c)-(a+b+c)=80a+8b=2(40a+4b)=2*214=428.

Given: Quadratic Equation with variables a, b, c & x [Also p] Equation answering the quadratic given.

To Do: Simplify the equation p(7) - p(3) = 214 p(7-3) = 214 p(4) = 214 p = 214 / 4 p = 53.5 Now, Substitute the value of 'p' in the the next equation.

p(9) - p(1) = ? p(9 -1) = ? p(8) = ?

Substituting the value of 'p'

53.5 * 8 = 428

We can write $p(7)-p(3)=214$ as $49a+7b+c-(9a+3b+c)=214$ . Simplifying, we get $40a+4b=214$ . Now let's work with $p(9)-p(1)$ . We can write $p(9)-p(1)$ as $81a+9b+c-(a+b+c)$ Simplifying, we get $80a+8b$ . $80a+8b$ is just $2\cdot(40a+4b)$ , and we can substitute 214 into $40a+4b$ , so $80a+8b=\boxed{428}$

By inputting 7 and 3 as $x$ we get:

$(49a + 7b + c) - (9a + 3b + c)=(40a + 4b)$

By also inputting 9 and 1 we get:

$(81a + 9b + c) - (a + b + c)=(80a + 8b)$

This means that:

$p(9) -p(1)=2*(p(7)-p(3))$

Since:

$p(7)-p(3)=214$

$p(9)-p(1)=428$

If p(7) - p(3) = 214 then

(49a + 7b + c) - (9a + 3b +c) = 214

40a + 4b = 214

p(9) - p(1) can be defined by the following expression which we'll call x

(81a +9b + c) - (a + b + c) = x

80a + 8b = x

Factoring we get

8(10a + b) = x

Factoring 40a + 4b we get 4(10a + b) = 214

Because 214/(10a + b) = 4 and x/(10a + b) = 8 we know that x must be twice as large as 214 so 428 is our answer

P(7) = 49a+7b+c
P(3) = 9a+3b+c
P(9) =81a+9b+c
P(1) = a+b+c
P(7)-P(3) = 49a+7b+c-9a-3b-c
= 40a+4b
therefore, 40a+4b=214 ----------------------(1)
P(9)-P(1) = 81a+9b+c-a-b-c
= 80a+8b
therefore, 80a+8b = k ----------------------(2)
Now, we have a system of equation to solve from (1) and (2)
40a+4b=214
80a+8b=k -------> divide by 2 --------> 40a+4b=k/2
So, 214 = k/2 -------------> k = 428

1 p(7) - p(3) = 214

p(4) = 214

1# p(9) - p(1) = p(8) p(8) is two times the amount of p(4) or 214. 214 * 2 = 428

P(7)-P(3)= 49a+7b+c-9a-3b-c=40a+4b=214

P(9)-P(1)=81a+9b+c-a-b-c=80a+8b=2(40a+4b)=2*214=428

p(x) = a $x^2$ + bx + c

as we may know in the first case p(7) , x is equal to 7

substitute the value of x, so we obtain these equations

p(7) = 49a+7b+c

p(3) = 9a+3b+c

and then subtract p(3) from p(7),so we obtain that 40a+4b = 214

for the p(9) and p(1)

p(9) = 81a+9b+c

p(1) = a+b+c

then subtract p(1) from p(9) ,we will then obtain 80a+8b

if 40a+4b = 214

80a+8b can be simplified to this form 2(40a+4b)

that is the same to 2(214)

so the value of 80a+80b is $\boxed{428}$

p(7)-p(3)= 49a+7b+c - (9a+3b+c) = 40a+4b=214 (I)
p(9)-p(1)= 81a+9b+c - (a+b+c) = 80a+ 8b (II)
So, it´s easy to see that if you multiply (I) by 2, it´s (II).
Finally you can find that 80a+8b = 428