Answer:
a) x² +9
b) x⁴ + 2 x² + 1
Step-by-step explanation:
a) (x+3 i)(x− 3 i)
= x² - (3 i ) x + (3 i ) x - (3 i)²
= x² - 9 i² ∵ i² = -1
= x² - 9 (-1)
= x² +9
b) (x+i)²(x-i)²
= (x² + i² + 2 x i)(x² + i² - 2 x i)
= ((x² -1 )+ 2 x i)((x² -1) - 2 x i)
using identity
(a + b)(a - b) = a² - b²
= (x² -1)² - (2 x i)²
= x⁴ + 1 - 2x² - 4x² i²
= x⁴ + 1 - 2 x² + 4 x²
= x⁴ + 2 x² + 1
The expressions in standard form are x²+9 and x⁴+x²+x³i-xi+1
Step-by-step explanation:
The question requires you to write (x+3i)(x− 3i) in standard form
(x+3i)(x− 3i)-----------------------------distribute
x(x-3i)+3i(x-3i)----------------------open brackets
x*x-x*3i+3i*x-3i*3i
x²-3ix+3ix--9
x²+9
b) (x+i)²(x-i)²
(x+i)²= x(x+i)+i(x+i)
(x+i)²=x²+xi+xi+i²
(x+i)²=x²+2xi-1
and
(x-i)²= (x-i)(x-i)
(x-1)²=x(x-i)-i(x-i)
(x-1)²=x²-xi-xi-1
(x-1)²=x²-2xi-1
Then
(x²+2xi-1)(x²-2xi-1) will be
x²(x²-xi-1)+2xi(x²-2xi-1)-1(x²-xi-1)
x⁴-x³i-x²+2x³i+4x²-2xi - x²+xi+1
x⁴-x²-x²-x³i+2x³i+4x²-x²-2xi+xi+1
x⁴-2x²+x³i+3x²-xi+1
x⁴+x²+x³i-xi+1
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Keywords: standard form, expressions, imaginary numbers
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