Find Factors Of Polynomial F(x) = 6x^4 - 21x^3 - 4x^2 + 24x - 35

Hey guys, let's dive into a super interesting math problem today! We're going to figure out which of the given options is a factor of the polynomial $f(x) = 6x^4 - 21x^3 - 4x^2 + 24x - 35$. Finding factors of polynomials can seem a bit tricky at first, but with the right approach, it's totally manageable. We've got four suspects: A. $2x - 7$, B. $2x + 7$, C. $3x - 7$, and D. $3x + 7$. Our mission is to test each one and see which one divides $f(x)$ perfectly, meaning there's no remainder.

This is where the Factor Theorem comes in handy, and trust me, it's a game-changer for problems like this. The Factor Theorem states that for a polynomial $f(x)$, a linear expression $(ax - b)$ is a factor if and only if $f(b/a) = 0$. In simpler terms, if we plug the root of the potential factor into the polynomial and get zero, then that linear expression is indeed a factor. So, for each option, we'll find its root and then substitute it into our polynomial $f(x)$. Let's get started!

Testing Option A: $2x - 7$

First up, we have option A, which is $2x - 7$. To find the root of this expression, we set it equal to zero: $2x - 7 = 0$. Solving for $x$, we get $2x = 7$, which means $x = 7/2$. Now, we need to substitute this value of $x$ into our polynomial $f(x) = 6x^4 - 21x^3 - 4x^2 + 24x - 35$. This might look a bit daunting with fractions, but we'll take it step-by-step.

$f(7/2) = 6(7/2)^4 - 21(7/2)^3 - 4(7/2)^2 + 24(7/2) - 35$

Let's calculate each term:

$(7/2)^2 = 49/4$ $(7/2)^3 = 343/8$ $(7/2)^4 = 2401/16$

Now, substitute these back:

$f(7/2) = 6(2401/16) - 21(343/8) - 4(49/4) + 24(7/2) - 35$

Simplify the multiplications:

$6(2401/16) = (3 imes 2)(2401)/(8 imes 2) = 3(2401)/8 = 7203/8$ $21(343/8) = 7203/8$ $4(49/4) = 49$ $24(7/2) = 12 imes 7 = 84$

So, $f(7/2) = 7203/8 - 7203/8 - 49 + 84 - 35$

$f(7/2) = 0 - 49 + 84 - 35$

$f(7/2) = -49 + 49$

$f(7/2) = 0$

Boom! We got zero! According to the Factor Theorem, since $f(7/2) = 0$, then $(2x - 7)$ is a factor of the polynomial $f(x)$. Since we found our answer, we could technically stop here. But for the sake of practice and to be absolutely sure, let's quickly check the other options to see what happens.

Testing Option B: $2x + 7$

For option B, $2x + 7$, we set it to zero: $2x + 7 = 0$. This gives us $2x = -7$, so $x = -7/2$. Now, let's plug this into $f(x)$:

$f(-7/2) = 6(-7/2)^4 - 21(-7/2)^3 - 4(-7/2)^2 + 24(-7/2) - 35$

Notice that the terms with odd powers will result in negative values, while even powers will be positive. Let's calculate:

$(-7/2)^2 = 49/4$ $(-7/2)^3 = -343/8$ $(-7/2)^4 = 2401/16$

Substitute these back:

$f(-7/2) = 6(2401/16) - 21(-343/8) - 4(49/4) + 24(-7/2) - 35$

Simplify:

$6(2401/16) = 7203/8$ $-21(-343/8) = 7203/8$ $-4(49/4) = -49$ $24(-7/2) = 12 imes (-7) = -84$

So, $f(-7/2) = 7203/8 + 7203/8 - 49 - 84 - 35$

$f(-7/2) = 14406/8 - 168$

$f(-7/2) = 7203/4 - 168$

This clearly isn't zero, so $(2x + 7)$ is not a factor. See how the signs change things drastically?

Testing Option C: $3x - 7$

Moving on to option C, $3x - 7$. Setting it to zero, we get $3x = 7$, so $x = 7/3$. Let's plug this into $f(x)$:

$f(7/3) = 6(7/3)^4 - 21(7/3)^3 - 4(7/3)^2 + 24(7/3) - 35$

Let's calculate the powers of $7/3$:

$(7/3)^2 = 49/9$ $(7/3)^3 = 343/27$ $(7/3)^4 = 2401/81$

Now, substitute and simplify:

$f(7/3) = 6(2401/81) - 21(343/27) - 4(49/9) + 24(7/3) - 35$

$6(2401/81) = (2 imes 3)(2401)/(27 imes 3) = 2(2401)/27 = 4802/27$ $21(343/27) = (7 imes 3)(343)/(9 imes 3) = 7(343)/9 = 2401/9$ $4(49/9) = 196/9$ $24(7/3) = 8 imes 7 = 56$

So, $f(7/3) = 4802/27 - 2401/9 - 196/9 + 56 - 35$

To combine the fractions, we need a common denominator, which is 27:

$2401/9 = (2401 imes 3) / (9 imes 3) = 7203/27$ $196/9 = (196 imes 3) / (9 imes 3) = 588/27$

$f(7/3) = 4802/27 - 7203/27 - 588/27 + (56 - 35)$

$f(7/3) = (4802 - 7203 - 588)/27 + 21$

$f(7/3) = (-2401 - 588)/27 + 21$

$f(7/3) = -2989/27 + 21$

This is definitely not zero. So, $(3x - 7)$ is not a factor.

Testing Option D: $3x + 7$

Finally, let's test option D, $3x + 7$. Setting it to zero, we get $3x = -7$, so $x = -7/3$. Let's plug this into $f(x)$:

$f(-7/3) = 6(-7/3)^4 - 21(-7/3)^3 - 4(-7/3)^2 + 24(-7/3) - 35$

Let's calculate the powers of $-7/3$:

$(-7/3)^2 = 49/9$ $(-7/3)^3 = -343/27$ $(-7/3)^4 = 2401/81$

Substitute and simplify:

$f(-7/3) = 6(2401/81) - 21(-343/27) - 4(49/9) + 24(-7/3) - 35$

$6(2401/81) = 4802/27$ $-21(-343/27) = 7203/27$ $-4(49/9) = -196/9$ $24(-7/3) = 8 imes (-7) = -56$

So, $f(-7/3) = 4802/27 + 7203/27 - 196/9 - 56 - 35$

Use a common denominator of 27 for the fractions:

$-196/9 = (-196 imes 3) / (9 imes 3) = -588/27$

$f(-7/3) = (4802 + 7203 - 588)/27 - (56 + 35)$

$f(-7/3) = (12005 - 588)/27 - 91$

$f(-7/3) = 11417/27 - 91$

Again, this is definitely not zero. So, $(3x + 7)$ is not a factor.

Conclusion: The Winning Factor!

So, after diligently testing all four options using the Factor Theorem, we found that only option A, $2x - 7$, resulted in $f(x) = 0$ when its root was substituted. This means that $2x - 7$ is the only factor of the polynomial $f(x) = 6x^4 - 21x^3 - 4x^2 + 24x - 35$ among the choices provided.

It's pretty cool how the Factor Theorem simplifies these problems, right? Even though the numbers can get a bit messy with fractions, the logic is straightforward. Keep practicing, and you'll be factoring polynomials like a pro in no time! Let me know if you guys have any other math puzzles you want to tackle!