Question

Can someone please help me with no.9 I have test tomorrow

Answer 1

Given the expression:

`3x^4+5x^3+px^2+qx+6`

Factors of the expression are:

(x - 2) and (x +3).

Let's find the values of p and q.

To solve for p and q, take the following steps:

Step 1:

Equate the factors to zero and solve for x.

x - 2 = 0

Add 2 to both sides:

x - 2 + 2 = 0 + 2

x = 2

x + 3 = 0

Subtract 3 from both sides:

x+ 3 - 3 = 0 - 3

x = -3

Thus, we have:

x = 2 and -3

Step 2:

Now, solve the expression for f(2) and f(-3).

We have:

`\begin{gathered} f(2)=3(2)^4+5(2)^3+p(2)^2+q(2)+6 \\ \\ f(2)=48+40+4q+2q+6 \\ \text{Combine like terms:} \\ f(2)=48+40+6+4p+2q \\ \\ f(2)=94+4p+2q \\ \\ \text{Equate to zero:} \\ 94+4p+2q=0 \\ \\ 4p-2q=-94 \end{gathered}`

Solve for f(-3):

`\begin{gathered} f(-3)=3(-3)^4+5(-3)^3+p(-3)^2+q(-3)+6 \\ \\ f(-3)=243-135+9p-3q+6 \\ \\ f(-3)=9p-3q+114 \\ \\ 9p-3q=-114 \end{gathered}`

Now, we have the set of equations:

4p - 2q = -94..................equation 1

9p - 3q = - 114..................equation 2

Step 3:

Solve the set of equations simultaneously using the elimination method.

4p - 2q = -94

9p - 3q = -114

Multilply all terms in equation 1 by -3 and equation two by 2:

4p(-3) - 2q(-3) = -94(-3)

9p(2) - 3q(2) = -114(2)

-12p + 6q = 282................equation 3

18p - 6q = -228,...............equation 4

Add both equation 3 and 4

-12p + 6q = 282

+ 18p - 6q = -228

--------------------------------

6p + 0 = 54

6p = 54

Divide both sides by 6:

`\begin{gathered} (6p)/(6)=(54)/(6) \\ \\ p=9 \end{gathered}`

Substitute 9 for p in equation 1:

4p - 2q = -94

4(9) - 2q = -94

36 - 2q = -94

Subtract 36 from both sides:

36 - 36 - 2q = -94 - 36

-2q = -130

Divide both sides by -2:

`\begin{gathered} (-2q)/(-2)=(-130)/(-2) \\ \\ q=65 \end{gathered}`

Therefore, the values of p and q are:

p = 9

q = 65

ANSWER:

p = 9 and q = 65