Polynomial Division And Factor Theorem Problems

Hey guys! Let's dive into some cool math problems involving polynomial division and the factor theorem. We'll break down each problem step-by-step, so you can totally get it.

1. Finding the Quotient and Remainder of $x^4 - 2x^2 + 10x - 5$ divided by $x^2 - 2x - 3$

Alright, so the first question asks us to find the quotient and remainder when we divide the polynomial $x^4 - 2x^2 + 10x - 5$ by $x^2 - 2x - 3$. Polynomial division can seem intimidating, but it's just like long division with numbers, but with variables! Let’s tackle this step by step.

First off, remember your main goal: we're trying to figure out how many times $x^2 - 2x - 3$ fits into $x^4 - 2x^2 + 10x - 5$, and what’s left over after we take out as many whole chunks as possible. This leftover bit is our remainder.

Let's set up the long division. Write $x^4 - 2x^2 + 10x - 5$ inside the division bracket and $x^2 - 2x - 3$ outside. Now, we look at the highest power of $x$ in both polynomials. We ask ourselves, “What do we need to multiply $x^2$ by to get $x^4$?” The answer is $x^2$. So, we write $x^2$ above the division bracket, aligned with the $x^2$ term.

Next, we multiply the entire divisor ($x^2 - 2x - 3$) by $x^2$, which gives us $x^4 - 2x^3 - 3x^2$. We write this below the dividend ($x^4 - 2x^2 + 10x - 5$) and subtract. Make sure you line up like terms! Subtracting, we get $(x^4 - 2x^2 + 10x - 5) - (x^4 - 2x^3 - 3x^2) = 2x^3 + x^2 + 10x - 5$.

Now, we bring down the next term, which is $-5$. We repeat the process. What do we need to multiply $x^2$ by to get $2x^3$? The answer is $2x$. So, we write $+2x$ next to $x^2$ above the division bracket. Multiply the divisor ($x^2 - 2x - 3$) by $2x$, which gives us $2x^3 - 4x^2 - 6x$. Subtract this from $2x^3 + x^2 + 10x - 5$ to get $(2x^3 + x^2 + 10x - 5) - (2x^3 - 4x^2 - 6x) = 5x^2 + 16x - 5$.

One last time! What do we need to multiply $x^2$ by to get $5x^2$? The answer is $5$. So, we write $+5$ next to $+2x$ above the division bracket. Multiply the divisor ($x^2 - 2x - 3$) by $5$, which gives us $5x^2 - 10x - 15$. Subtract this from $5x^2 + 16x - 5$ to get $(5x^2 + 16x - 5) - (5x^2 - 10x - 15) = 26x + 10$.

Since the degree of $26x + 10$ (which is 1) is less than the degree of $x^2 - 2x - 3$ (which is 2), we can't divide any further. This means $26x + 10$ is our remainder. So, the quotient is $x^2 + 2x + 5$ and the remainder is $26x + 10$.

2. Finding the Quotient and Remainder of $2x^3 + 3x^2 - 32x + 15$ divided by $2x - 1$

Next up, we need to divide $2x^3 + 3x^2 - 32x + 15$ by $2x - 1$. Same drill as before, just different numbers!

We're looking to see how many times $2x - 1$ fits nicely into $2x^3 + 3x^2 - 32x + 15$, and what, if anything, is left over.

Let's set up the long division again. Write $2x^3 + 3x^2 - 32x + 15$ inside the bracket and $2x - 1$ outside. Now, what do we need to multiply $2x$ by to get $2x^3$? The answer is $x^2$. Write $x^2$ above the division bracket, aligned with the $x^2$ term.

Multiply the divisor ($2x - 1$) by $x^2$, which gives us $2x^3 - x^2$. Write this below the dividend and subtract: $(2x^3 + 3x^2 - 32x + 15) - (2x^3 - x^2) = 4x^2 - 32x + 15$.

Bring down the next term, $+15$. What do we need to multiply $2x$ by to get $4x^2$? The answer is $2x$. So, we write $+2x$ next to $x^2$ above the division bracket. Multiply the divisor ($2x - 1$) by $2x$, which gives us $4x^2 - 2x$. Subtract this from $4x^2 - 32x + 15$ to get $(4x^2 - 32x + 15) - (4x^2 - 2x) = -30x + 15$.

One more time! What do we need to multiply $2x$ by to get $-30x$? The answer is $-15$. So, we write $-15$ next to $+2x$ above the division bracket. Multiply the divisor ($2x - 1$) by $-15$, which gives us $-30x + 15$. Subtract this from $-30x + 15$ to get $(-30x + 15) - (-30x + 15) = 0$.

Since we got a remainder of 0, that means $2x - 1$ divides perfectly into $2x^3 + 3x^2 - 32x + 15$. The quotient is $x^2 + 2x - 15$ and the remainder is $0$.

3. Finding the Value of P and Other Factors of $g(x) = -4x^3 + 2x^2 + Px - 2$ given $x + 1$ is a factor

Okay, this time, we're dealing with the factor theorem. The factor theorem basically says that if $(x - c)$ is a factor of a polynomial $g(x)$, then $g(c) = 0$. In simpler terms, if you plug in the value that makes the factor equal to zero, the whole polynomial will equal zero.

In our case, we know that $(x + 1)$ is a factor of $g(x) = -4x^3 + 2x^2 + Px - 2$. This means that if we plug in $x = -1$ into $g(x)$, the result should be zero. Let’s do it!

So, $g(-1) = -4(-1)^3 + 2(-1)^2 + P(-1) - 2 = 0$. Simplifying, we get $-4(-1) + 2(1) - P - 2 = 0$, which becomes $4 + 2 - P - 2 = 0$. This simplifies further to $4 - P = 0$. Solving for $P$, we find that $P = 4$.

Now that we know $P = 4$, our polynomial is $g(x) = -4x^3 + 2x^2 + 4x - 2$. We know $(x + 1)$ is a factor, so we can divide $g(x)$ by $(x + 1)$ to find the other factors. Let's use synthetic division for this. Set up the synthetic division with $-1$ (the root of $x + 1$) and the coefficients of $g(x)$: $-4, 2, 4, -2$.

Bring down the $-4$. Multiply $-1$ by $-4$ to get $4$, and add it to $2$ to get $6$. Multiply $-1$ by $6$ to get $-6$, and add it to $4$ to get $-2$. Multiply $-1$ by $-2$ to get $2$, and add it to $-2$ to get $0$. The remainder is 0, which confirms that $(x + 1)$ is indeed a factor.

The result of the synthetic division gives us the coefficients of the quotient, which is a quadratic: $-4x^2 + 6x - 2$. We can simplify this by dividing by $-2$ to get $2x^2 - 3x + 1$. Now, let's factor this quadratic.

We're looking for two numbers that multiply to $2$ and add up to $-3$. Those numbers are $-1$ and $-2$. So, we can factor $2x^2 - 3x + 1$ as $(2x - 1)(x - 1)$.

Therefore, the factors of $g(x) = -4x^3 + 2x^2 + 4x - 2$ are $(x + 1)$, $(2x - 1)$, and $(x - 1)$.

And there you have it! We've tackled polynomial division, the factor theorem, and synthetic division. Keep practicing, and you'll become a polynomial pro in no time!