Asked by Juan

Find m(5) using direct subsitution m(x)=x+x^2-1

Answers

Answered by Juan
Find w(-3) using synthetic substitution w(x)= 11x^3-6x^2+2
Answered by Juan
(X^4+6x^3-x^2-5x+1) divide (x-2) use synthetic division if possible
Answered by Juan
List the possible zeros for the polynomial
H(x)=2x^3+5x^2-31x-15
Answered by Bot
To find m(5), we simply substitute 5 for x in the expression for m(x):

m(5) = 5 + 5^2 - 1

m(5) = 5 + 25 - 1

m(5) = 29

Therefore, m(5) = 29.
Answered by Bot
To find w(-3) using synthetic substitution, we follow these steps:

Step 1: Write the coefficients of the polynomial in a row and leave some space for the calculations. In this case, we have:

11 -6 0 2

Step 2: Bring down the first coefficient:

11

Step 3: Multiply by the given value of x and write the result below the next coefficient:

11 -6
-33

Step 4: Add the numbers in the second row and write the result below the line:

11 -6
-33 0

Step 5: Repeat steps 3 and 4 for the next two coefficients:

11 -6 0
-33 9
99
-------------
11 3 99

Step 6: The result is the last number in the row, which corresponds to the constant term of the polynomial:

w(-3) = 11(-3)^3 - 6(-3)^2 + 2 = -297 + 54 + 2 = -241

Therefore, w(-3)=-241.
Answered by Bot
We can use synthetic division to divide the polynomial x^4 + 6x^3 - x^2 - 5x + 1 by x - 2:

```
2 | 1 6 -1 -5 1
| 2 16 30 50
|----------------------------
| 1 8 15 25 51
```

Therefore,

x^4 + 6x^3 - x^2 - 5x + 1 = (x - 2)(x^3 + 8x^2 + 15x + 25) + 51

So, the quotient is x^3 + 8x^2 + 15x + 25 with a remainder of 51.
Answered by Bot
We can use the Rational Root Theorem to list the possible zeros of the polynomial H(x) = 2x^3 + 5x^2 - 31x - 15.

The Rational Root Theorem states that all rational roots of a polynomial with integer coefficients must have the form of p/q, where p is a factor of the constant term and q is a factor of the leading coefficient. In this case, the constant term is -15 and the leading coefficient is 2.

The factors of the constant term -15 are ±1, ±3, ±5, and ±15. The factors of the leading coefficient 2 are ±1 and ±2. Therefore, the possible rational zeros of the polynomial H(x) are:

±1/2, ±1, ±3/2, ±3, ±5/2, ±5, ±15/2, ±15.

Note that this list includes both positive and negative values. However, it's important to keep in mind that not all of these possible zeros are necessarily actual zeros of the polynomial.

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