Generate an image of a vibrant, multi-colored graph that represents a polynomial function. The graph should be drawn with clear intersecting points at two different values of 'x'. In addition, incorporate a thought bubble into the image that contains a numeric value, representing the concept of a remainder.

1.) when the expression 4x^2-3x-8 is divided by x-a, the remainder is 2. find the value of a.

2.) the polynomial 3x^3+mx^2+nx+5 leaves a remainder of 128 when divided by x-3 and a remainder of 4 when divided by x+1. calculate the remainder when the polynomial is divided by x-3.

tnx...

let f(x) = 4x^2 - 3x - 8 and use the Remainder Theorem,

f(a) = 4a^2 - 3a - 8 = 2
4a^2 - 3a - 10 = 0
(a-2)(4a + 5) = 0
a = 2 or a = -5/4

the 2nd is done the same way:
f(3) = 81 + 9m + 3n + 5 = 128
9m + 3n = 42
3m + n = 14

f(-1) = -3 - m - n + 5 = 4
-m-n = 2

add them:
2m = 16
m = 8 , then 24 + n = 14 ----> n = -10

so f(x) = 3x^3 + 8x^2 -10x + 5
f(3) = 81 + 72 - 30 + 5 = 128
DUH, wasn't that given ?

thanks... :)

What is P(3) if P(x)=5x^4 + 7x^2 - 1? *

1.) Well, if the remainder is 2, then we can say that nobody really likes the number 2 and it always has to stick around, even when we're dividing polynomials. So, let's divide 4x^2-3x-8 by x-a and see what happens.

Since the remainder is 2, we have:

4a^2 - 3a - 8 = 2

Now, don't panic! We just need to solve this quadratic equation to find the value of a. Trust me, math can be funny sometimes!

By rearranging the terms and adding -2 to both sides, we get:

4a^2 - 3a - 10 = 0

Now, we can use the quadratic formula to find the value of a. But let's be cautious, quadratic formulas can be a bit explosive — you never know what kind of numbers they will spit out!

Using the quadratic formula: a = (-b ± √(b^2 - 4ac)) / 2a

For our equation, a = (-(-3) ± √((-3)^2 - 4(4)(-10))) / (2 * 4)

Simplifying this expression, we find:

a = (3 ± √(9 + 160)) / 8

a = (3 ± √(169)) / 8

Now, remember, when it comes to quadratic formulas, there are always two answers, like trying to decide between chocolate or vanilla ice cream. So, let's consider both possibilities:

a1 = (3 + 13) / 8 = 16/8 = 2

a2 = (3 - 13) / 8 = -10/8 = -5/4

So, the possible values for a are 2 and -5/4. I hope these answers don't make you feel divided!

2.) Ah, the funny business continues! So, we have this polynomial 3x^3 + mx^2 + nx + 5, and it leaves a remainder of 128 when divided by x-3, and a remainder of 4 when divided by x+1. That polynomial must be feeling quite remainder-ed!

To find the remainder when the polynomial is divided by x-3, we need to use the remainder theorem. According to this high-tech theorem, if we substitute the divisor (x-3) into the polynomial and get the remainder, that remainder will be our answer.

So let’s plug x=3 into our polynomial:

3(3)^3 + m(3)^2 + n(3) + 5 = 128

Now, let me do some calculations for you. Give me a second...

27 + 9m + 3n + 5 = 128

9m + 3n = 128 - 27 - 5

9m + 3n = 96

Now comes the fun part where you solve this equation and get the values for m and n! I'm not a mind reader, but I believe in your mathematical abilities.

1.) To find the value of 'a' when the expression 4x^2-3x-8 is divided by x-a and the remainder is 2, we can use the Remainder Theorem.

According to the Remainder Theorem, when a polynomial is divided by a linear factor (x-a), the remainder is equal to the polynomial evaluated at 'a'.

So, in this case, we have:
4a^2 - 3a - 8 = 2

To solve this equation, we rearrange it:
4a^2 - 3a - 10 = 0

Now, we can factor or use the quadratic formula to solve for 'a'. Factoring might be a bit challenging here, so let's use the quadratic formula.

The quadratic formula states that the solutions for the equation ax^2 + bx + c = 0 are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation 4a^2 - 3a - 10 = 0, we have:
a = (-(-3) ± √((-3)^2 - 4*4*(-10))) / (2 * 4)
a = (3 ± √(9 + 160)) / 8
a = (3 ± √169) / 8
a = (3 ± 13) / 8

Therefore, the two possible values for 'a' are:
a1 = (3 + 13) / 8 = 16/8 = 2
a2 = (3 - 13) / 8 = -10/8 = -5/4

So, the value of 'a' can be either 2 or -5/4.

2.) To calculate the remainder when the polynomial 3x^3 + mx^2 + nx + 5 is divided by x-3, we are given that the remainder is 128.

Again, we can use the Remainder Theorem to solve this.

When the polynomial is divided by x-3, the remainder is equal to the polynomial evaluated at 3, which gives us:
3(3)^3 + m(3)^2 + n(3) + 5 = 128

Simplifying this equation, we get:
81 + 9m + 3n + 5 = 128
86 + 9m + 3n = 128

To find the remainder, we only need the constant term, which is 86.

So, the remainder when the polynomial is divided by x-3 is 86.

Stop giving wrong anwser bro you made me fail