Question

1.) What are the zeros of the polynomial? f(x)=x^4-x^3-16x^2+4x+48.

Answer 1

Answer:

3.) $\displaystyle [x - 2][x^2 + 2][x + 4]$

2.) $\displaystyle 2\:complex\:solutions → x^2 + 3x + 6 >> -\frac{3 - i\sqrt{15}}{2}, -\frac{3 + i\sqrt{15}}{2}$

1.) $\displaystyle 4, -3, 2, and\:-2$

Step-by-step explanation:

3.) By the Rational Root Theorem, we would take the Least Common Divisor [LCD] between the leading coefficient of 1, and the initial value of −16, which is 1, but we will take 2 since it is the fourth root of 16; so this automatically makes our first factor of $\displaystyle x - 2.$Next, since the factor\divisor is in the form of $\displaystyle x - c$, use what is called Synthetic Division. Remember, in this formula, −c gives you the OPPOSITE terms of what they really are, so do not forget it. Anyway, here is how it is done:

2| 1 2 −6 4 −16

↓ 2 8 4 16

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1 4 2 8 0 → $\displaystyle x^3 + 4x^2 + 2x + 8$

You start by placing the c in the top left corner, then list all the coefficients of your dividend [x⁴ + 2x³ - 6x² + 4x - 16]. You bring down the original term closest to c then begin your multiplication. Now depending on what symbol your result is tells you whether the next step is to subtract or add, then you continue this process starting with multiplication all the way up until you reach the end. Now, when the last term is 0, that means you have no remainder. Finally, your quotient is one degree less than your dividend, so that 1 in your quotient can be an x³, the 4x² follows right behind it, bringing 2x right up against it, and bringing up the rear, 8, giving you the quotient of $\displaystyle x^3 + 4x^2 + 2x + 8.$

However, we are not finished yet. This is our first quotient. The next step, while still using the Rational Root Theorem with our first quotient, is to take the Least Common Divisor [LCD] of the leading coefficient of 1, and the initial value of 8, which is −4, so this makes our next factor of $\displaystyle x + 4.$Then again, we use Synthetic Division because $\displaystyle x + 4$is in the form of $\displaystyle x - c$:

−4| 1 4 2 8

↓ −4 0 −8

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1 0 2 0 → $\displaystyle x^2 + 2$

So altogether, we have our four factors of $\displaystyle [x^2 + 2][x + 4][x - 2].$

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2.) By the Rational Root Theorem again, this time, we will take −1, since the leading coefficient & variable\degree and the initial value do not share any common divisors other than the special number of 1, and it does not matter which integer of 1 you take first. This gives a factor of $\displaystyle x + 1.$Then start up Synthetic Division again:

−1| 1 3 5 −3 −6

↓ −1 −2 −3 6

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1 2 3 −6 0 → $\displaystyle x^3 + 2x^2 + 3x - 6$

Now we take the other integer of 1 to get the other factor of $\displaystyle x - 1,$then repeat the process of Synthetic Division:

1| 1 2 3 −6

↓ 1 3 6

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1 3 6 0 → $\displaystyle x^2 + 3x + 6$

So altogether, we have our three factors of $\displaystyle [x - 1][x^2 + 3x + 6][x + 1].$

Hold it now! Notice that $\displaystyle x^2 + 3x + 6$is unfactorable. Therefore, we have to apply the Quadratic Formula to get our two complex solutions, $\displaystyle a + bi$[or zeros in this matter]:

$\displaystyle -b ± \frac{\sqrt{b^2 - 4ac}}{2a} = x \\ \\ -3 ± \frac{\sqrt{3^2 - 4[1][6]}}{2[1]} = x \\ \\ -3 ± \frac{\sqrt{9 - 24}}{2} = x \\ \\ -3 ± \frac{\sqrt{-15}}{2} = x \\ \\ -3 ± i\frac{\sqrt{15}}{2} = x \\ \\ -\frac{3 - i\sqrt{15}}{2}, -\frac{3 + i\sqrt{15}}{2} = x$

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1.) By the Rational Root Theorem one more time, this time, we will take 4 since the initial value is 48 and that 4 is the root of the polynomial. This gives our automatic factor of $\displaystyle x - 4.$Then start up Synthetic Division again:

4| 1 −1 −16 4 48

↓ 4 12 −16 −48

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1 3 −4 −12 0 → $\displaystyle x^3 + 3x^2 - 4x - 12$

We can then take −3 since it is a root of this polynomial, giving us the factor of $\displaystyle x + 3:$

−3| 1 3 −4 −12

↓ −3 0 12

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1 0 −4 0 → $\displaystyle x^2 - 4 >> [x - 2][x + 2]$

So altogether, we have our four factors of $\displaystyle [x - 2][x + 3][x + 2][x - 4],$and when set to equal zero, you will get $\displaystyle 4, -3, 2, and\:-2.$

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