Polynomial Root Justification And Coefficient Determination

Let's dive into this polynomial problem together! We've got a cubic polynomial, $P(x) = -56x^3 + 325x^2 + 81x - 90$, and we need to figure out why 6 is a root and then find some coefficients. It sounds like a fun challenge, so let's get started!

Justifying That 6 Is a Root of P(x)

So, the first part of our mission is to show that 6 is indeed a root of the polynomial $P(x)$. What does it mean for a number to be a root, anyway? Well, simply put, a number 'r' is a root of a polynomial if plugging 'r' into the polynomial gives us zero. In other words, $P(r) = 0$. So, in our case, we need to show that $P(6) = 0$.

Let's get our hands dirty and actually substitute $x = 6$ into our polynomial: $P(6) = -56(6)^3 + 325(6)^2 + 81(6) - 90$. Now, we just need to carefully crunch these numbers.

First up, we have $6^3$, which is $6 * 6 * 6 = 216$. Next, we've got $6^2$, which is $6 * 6 = 36$. Now we can rewrite our expression as:

$P(6) = -56 * 216 + 325 * 36 + 81 * 6 - 90$

Time for some more multiplication! Let's calculate each term:

• $-56 * 216 = -12096$
• $325 * 36 = 11700$
• $81 * 6 = 486$

Now we can substitute these values back into our expression:

$P(6) = -12096 + 11700 + 486 - 90$

Let's add and subtract these numbers carefully. First, let's combine the negative terms: $-12096 - 90 = -12186$. Now, let's combine the positive terms: $11700 + 486 = 12186$. So, our expression becomes:

$P(6) = -12186 + 12186$

And what do you know? These cancel each other out perfectly:

$P(6) = 0$

Ta-da! We've shown that $P(6) = 0$, which means that 6 is indeed a root of the polynomial $P(x)$. We did it! The key here was simply substituting the value and carefully doing the arithmetic. Remember, always double-check your calculations to avoid silly mistakes.

In summary, to justify that 6 is a root, we directly substituted 6 into the polynomial P(x) and showed that the result is equal to zero. This confirms that 6 is a zero (or root) of the polynomial. This detailed step-by-step approach is crucial for demonstrating the reasoning and arriving at the correct conclusion.

Determining the Values of Real Numbers a, b, and c

Okay, guys, now that we've successfully proven that 6 is a root of our polynomial $P(x)$, the next part of the puzzle is to determine the values of the real numbers a, b, and c. But, what exactly are a, b, and c related to? Well, the problem implies that we should be able to express our polynomial $P(x)$ in a different form, one that involves these mysterious a, b, and c.

Since we know that 6 is a root of $P(x)$, this means that $(x - 6)$ must be a factor of $P(x)$. This is a crucial piece of information! Why? Because it allows us to rewrite $P(x)$ as a product of $(x - 6)$ and another polynomial. Since $P(x)$ is a cubic polynomial (degree 3), and we're factoring out a linear term $(x - 6)$ (degree 1), the other polynomial must be a quadratic (degree 2). So, we can write:

$P(x) = (x - 6)(ax^2 + bx + c)$

See? There are our a, b, and c! Our goal now is to find the specific values of these coefficients. How do we do that? Well, we know what $P(x)$ is explicitly: $P(x) = -56x^3 + 325x^2 + 81x - 90$. So, what if we expand the factored form $(x - 6)(ax^2 + bx + c)$ and then compare the coefficients with our original polynomial? That sounds like a plan!

Let's expand the factored form. We'll carefully distribute each term in $(x - 6)$ across the terms in $(ax^2 + bx + c)$:

$(x - 6)(ax^2 + bx + c) = x(ax^2 + bx + c) - 6(ax^2 + bx + c)$

Now, let's distribute further:

$= ax^3 + bx^2 + cx - 6ax^2 - 6bx - 6c$

Okay, let's group like terms together:

$= ax^3 + (b - 6a)x^2 + (c - 6b)x - 6c$

Great! Now we have an expanded form of $P(x)$ in terms of a, b, and c: $P(x) = ax^3 + (b - 6a)x^2 + (c - 6b)x - 6c$. And we also know the original form of $P(x)$: $P(x) = -56x^3 + 325x^2 + 81x - 90$. The magic happens when we equate the coefficients of corresponding terms. This gives us a system of equations:

• Coefficient of $x^3$: $a = -56$
• Coefficient of $x^2$: $b - 6a = 325$
• Coefficient of $x$: $c - 6b = 81$
• Constant term: $-6c = -90$

Look at that! We've transformed our polynomial problem into a system of linear equations. And even better, the first and last equations are super simple. Let's solve them first.

From the first equation, we immediately get $a = -56$. Awesome! From the last equation, $-6c = -90$, we can divide both sides by -6 to get $c = 15$. Fantastic! Now we have two of our unknowns. Let's use these values to find 'b'.

We have the equation $b - 6a = 325$. We know $a = -56$, so let's substitute:

$b - 6(-56) = 325$

$b + 336 = 325$

Subtract 336 from both sides:

$b = 325 - 336$

$b = -11$

And there we have it! We've found all the values: $a = -56$, $b = -11$, and $c = 15$. The key here was recognizing the factor $(x - 6)$, expanding the factored form, and then equating coefficients. This is a powerful technique for working with polynomials.

Therefore, we've determined the values of the real numbers a, b, and c such that the polynomial P(x) can be expressed as P(x) = (x - 6)(ax^2 + bx + c). We found that a = -56, b = -11, and c = 15 by expanding the factored form and equating coefficients with the original polynomial. This methodical approach allowed us to solve for the unknowns and complete the problem.

Conclusion

Guys, we've tackled a challenging polynomial problem and come out victorious! We successfully justified why 6 is a root of the given polynomial by direct substitution. Then, we masterfully determined the values of a, b, and c by leveraging the factor theorem, expanding the factored form, and solving the resulting system of equations. Remember, practice makes perfect, so keep working on these types of problems, and you'll become polynomial pros in no time! Remember the importance of careful calculation and methodical steps in solving such problems. These skills are essential not just in mathematics, but in many problem-solving scenarios in life. So keep practicing and keep learning!