To solve the problem, we need to find the values of a and b in the polynomial P(x) = 3x^4 + ax^3 + bx^2 + 25x - 10, given that it has a factor 3x^2 + 5x - 2. First, let's factor the quadratic 3x^2 + 5x - 2. We can factor it as:

3x^2 + 5x - 2 = (3x - 1)(x + 2)

This means that both 3x - 1 and x + 2 are factors of P(x).

Since 3x - 1 is a factor, P\left(\frac{1}{3}\right) = 0. Substituting x = \frac{1}{3} into P(x), we have :

P\left(\frac{1}{3}\right) = 3\left(\frac{1}{3}\right)^4 + a\left(\frac{1}{3}\right)^3 + b\left(\frac{1}{3}\right)^2 + 25\left(\frac{1}{3}\right) - 10 = 0

Simplifying each term:

\frac{3}{81} + \frac{a}{27} + \frac{b}{9} + \frac{25}{3} - 10 = 0

Multiply through by 27 to clear the fractions:

1+a+3b+225-270=0

Simplifying gives:

a + 3b = 44

Next, since x + 2 is a factor, P(-2) = 0. Substituting x = -2 into P(x), we have:

P(-2) = 3(-2)^4 + a(-2)^3 + b(-2)^2 + 25(-2) - 10 = 0

Simplifying each term:

48 - 8a + 4b - 50 - 10 = 0

This simplifies to:

-8a + 4b = 12

Dividing through by 4:

-2a + b = 3

Now, we have the system of equations:

\begin{cases} a + 3b = 44 \\ -2a + b = 3 \end{cases}

To solve this system, we can use substitution or elimination. Let's use elimination. Multiply the second equation by 3 to align the coefficients of b :

\begin{cases} a + 3b = 44 \\ -6a + 3b = 9 \end{cases}

Subtract the second equation from the first:

a + 3b - (-6a + 3b) = 44 - 9

This simplifies to:

7a = 35

Solving for a:

\boxed{a=5}

Substitute a = 5 back into the first equation:

5 + 3b = 44

Solving for b:

3b=39

Thus \boxed{b=13} .

Problem Bank - Polynomials

Problem Bank - Polynomials