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A

Anonymous

Nov 16, 2024

simplify (4-8i)(2-7i)?

I need help simplifying the following expressions:

  1. Simplify (4 - 8i)(2 - 7i).
  2. Simplify 27^(-2/3).
  3. Simplify i^314.
  4. Simplify (-8 + 8i) - (5 - 4i).
  5. Simplify (x^(2/3))^(4/5).

Thank you so much; I'm really lost on these!

8 Answers

A
Anonymous

Feb 01, 2025

(4-8i)(2-7i) = 8-56 – 16i – 28i = -48-44i

A
Anonymous

Jan 22, 2025

1. First multiply them together the same way you would multiply any two binomials, multiply 4 and 2, and 4 and -7i, then multiply -8i and 2, and -8i and -7i

8 – 28i – 16i + 56i^2

combine – 28i and -16i, and remember i = sq. rt. of -1, so i^2 = -1

8 – 44i – 56

combine 8 and -56

-48 – 44i

2. 27^-(2/3)

The negative exponent simply puts the number 27^(2/3) (no more negative exponent) as the denominator of a unit fraction (1 as the numerator)

so 1/(27^(2/3))

break up the 2/3 into (27^(1/3))^2

27^(1/3) = cube root of 27 = 3

so 3^2 = 9

so the whole thing put together is 1/9

3. i^31

since i = sq.rt.(-1)

i^2 = -1

going further, if you square that again, so i^4 (or (-1)^2) = 1

i^31 = i^28 * i^3 (since 28 is the highest multiple of 4, the first 28 i’s that you multiply together end up giving you 1.

so, 1 * i^3 = i^3 but i^2 = -1 so i^3 = -1i

-1i is the simplified answer.

4. just a subtraction/distribution problem, you have to distribute the minus sign,

-8 – 5 and 8i – (-4i)

-13 + 12i

5. (x^2/3)^(4/5)

and exponent of and exponent, means you can just multiply the exponents together,

2/3 * 4/5 = 8/15

so, x^(8/15)

A
Anonymous

Jan 25, 2025

1) Multiply the terms using F.O.I.L. (First, Outer, Inner, Last), and know that √-1 = i

( 4 – 8i ) ( 2 – 7i )

= 4×2 + 4x-7i + -8ix2 + -8i x -7i

= 8 – 28i -16i + 56i²

= 8 – 44i + 56(√-1)²

= 8 – 44i + 56x-1

= 8 – 44i – 56

= -48 – 44i

2) 27^-(2/3)

Anything to the power of a negative number is the same as one over the same number to the positive power:

= 1 / 27^(2/3)

= 1 / 27^(2 * ⅓)

= 1 / (27^2 ^ ⅓)

= 1 / (729 ^ ⅓)

Anything to the power of a fraction is a root. To the power ½ is a square root; to the power ⅓ is a cube root.

= 1 / ∛729

= 1 / 9

3) i=√-1

That means that i² = (√-1)² = -1

That means that i⁴= (√-1)⁴ = (-1)² = 1

i^31 = i^(4*7+2+1) = i^(4*7)i²i¹ = i^4^7 i² i¹ = 1^7 x -1 x i = 1 x -i = -i

4) Expand:

= -8 + 8i -5 + 4i

= -13 + 12i

5) If you raise a power to another power, you can simply multiply the exponents together:

x ^ 2/3 ^ 4/5

= x ^ [ (2*4)/(3*5) ]

= x ^ 8/15

A
Anonymous

Feb 15, 2025

just foil as you would normally, keeping in mind that i^2 = -1.

so:

(4 – 8 i ) (2 – 7 i )

firsts:

4 * 2 = 8

outer:

4 * -7 i = -28 i

inner:

2 * -8 i = -16 i

lasts :

– 8 i * – 7 i = 56 i ^2 = – 56.

so :

(4-8i)(2-7i)

8 – 16 i – 28 i – 56

= 8 – 44 i – 56

= – 48 – 44 i

J
Joe Hermiston

Jan 24, 2025

The FOIL technique. First, outdoors, interior, final. in simple terms like numerous different polynomial simplification. of direction i = ?(-a million). So it is going to look like this: First outdoors interior final (4)(5) + (4)(-7i) + (3i)(5) + (3i)(-7i) = 20 – 28i +15i – 21i² = 20 – 13i + 21 = 40-one – 13i (simplified)

A
Anonymous

Jan 23, 2025

(4 – 8i)(2 – 7i)

8 – 28i -16i – 56

64 – 44i

—————————–

2. 27^(1/3) = 3

27^(2/3) = 9

27^-(2/3) = 1/9

———————————–

i^31 = i

————————————

(-8 + 8i)-(5 -4i)

40 +32i -40i -+ 32

72 -8i

——————————

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