Asked by Gracie
Help please!
1. What are the real or imaginary solutions of the polynomial equation?
x^4-52x^2+576=0
A) 4,-4
B) 4,-6
C) 4,-4,6,-6
D) no solutions
2. What are the real or imaginary solutions of the polynomial equation?
x^3=216
A) -6,3+3i(sq root of 7), and 3-3i(sq root of 7)
B) -6, 3+3i(sq root of 3), and 3-3i(sq root of 3)
C) 6, 3+3i(sq root of 7), and 3-3i(sq root of 7)
D) 6, 3+3i(sq root of 3), and 3-3i(sq root of 3)
5. The dimensions, in inches, of a shipping box at W Ship 4 You can be expressed as width x, length x+5, and height 3x-1. The volume is about 7.6 ft^3. Find the dimension of the box in inches. Round to the nearest inch. (Hint: 1 ft^3=1,728 in.^3)
A) 15 in. by 20 in. by 44 in.
B) 12 in. by 17 in. by 35 in.
C) 15 in. by 20 in. by 45 in.
D) 12 in. by 17 in. by 36 in.
1. What are the real or imaginary solutions of the polynomial equation?
x^4-52x^2+576=0
A) 4,-4
B) 4,-6
C) 4,-4,6,-6
D) no solutions
2. What are the real or imaginary solutions of the polynomial equation?
x^3=216
A) -6,3+3i(sq root of 7), and 3-3i(sq root of 7)
B) -6, 3+3i(sq root of 3), and 3-3i(sq root of 3)
C) 6, 3+3i(sq root of 7), and 3-3i(sq root of 7)
D) 6, 3+3i(sq root of 3), and 3-3i(sq root of 3)
5. The dimensions, in inches, of a shipping box at W Ship 4 You can be expressed as width x, length x+5, and height 3x-1. The volume is about 7.6 ft^3. Find the dimension of the box in inches. Round to the nearest inch. (Hint: 1 ft^3=1,728 in.^3)
A) 15 in. by 20 in. by 44 in.
B) 12 in. by 17 in. by 35 in.
C) 15 in. by 20 in. by 45 in.
D) 12 in. by 17 in. by 36 in.
Answers
Answered by
oobleck
#1. x^4-52x^2+576 = 0
Easy enough to check. They even give you the values to try. Just do a few synthetic divisions to see whether the proposed values work. Synthetic division is fast and easy. Once you have found two roots, you end up with a quadratic, and that's no trouble.
However, I'd try C first, since you can just divide by x^2-16 or x^2-36 to check it.
Or, if you let u = x^2, you have a quadratic in u:
u^2 - 52u + 576 = (u-16)(u-36)
That means the factors are (x^2-16)(x^2-36)
#2. x^3 = 216 = 6^3
So, you know the real root is 6
The complex roots are 6cis(2π/3) and 6cis(4π/3)
or, 6(1±i√3)/2
#3. More of the same. You want to solve
x(x+5)(3x-1) = 7.6*1728
3x^3+14x^2-5x = 13132.8
3x^3 + 14x^2 - 5x - 13133 = 0
Now, you can go through the motions of trying to solve this, but a few seconds with your calculator shows that A gives a volume of 13200, which is pretty close. C and D are obviously wrong, since the dimensions do not fit the requirements.
Easy enough to check. They even give you the values to try. Just do a few synthetic divisions to see whether the proposed values work. Synthetic division is fast and easy. Once you have found two roots, you end up with a quadratic, and that's no trouble.
However, I'd try C first, since you can just divide by x^2-16 or x^2-36 to check it.
Or, if you let u = x^2, you have a quadratic in u:
u^2 - 52u + 576 = (u-16)(u-36)
That means the factors are (x^2-16)(x^2-36)
#2. x^3 = 216 = 6^3
So, you know the real root is 6
The complex roots are 6cis(2π/3) and 6cis(4π/3)
or, 6(1±i√3)/2
#3. More of the same. You want to solve
x(x+5)(3x-1) = 7.6*1728
3x^3+14x^2-5x = 13132.8
3x^3 + 14x^2 - 5x - 13133 = 0
Now, you can go through the motions of trying to solve this, but a few seconds with your calculator shows that A gives a volume of 13200, which is pretty close. C and D are obviously wrong, since the dimensions do not fit the requirements.
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