Question

Complete the following exercises by applying polynomial identities to complex numbers. Factor x2 + 64. Check your work. Factor 16x2 + 49. Check your work. Find the product of (x + 9i)2. Find the product of (x − 2i)2. Find the product of (x + (3+5i))2.

Answer 1

Answer:

Step-by-step explanation:

x^2+64=x^2-(8i)^2=(x-8i)(x+8i)

16x^2+49=(4x)^2-(7i)^2=(4x-7i)(4x+7i)

(x+9i)^2=x^2+2x\cdot9i+(9i)^2=x^2+18xi+81i^2=x^2+18xi-81=x^2-81+18xi

(x-2i)^2=x^2-2x\cdot2i+(2i)^2=x^2-4xi+4i^2=x^2-4xi-4=x^2-4-4xi

(x+(3+5i))^2=x^2+2x(3+5i)+(3+5i)^2\\\\=x^2+6x+10xi+3^2+2\cdot3\cdot5i+(5i)^2\\\\=x^2+6x+10xi+9+30i+25i^2=x^2+6x+10xi+9+30i-25\\\\=x^2+6x+10xi+30i-16=x^2+6x-16+(10x+30)i

Used:\\i^2=-1\\\\a^2-b^2=(a-b)(a+b)\\\\(a+b)^2=a^2+2ab+b^2

Answer 2

$x^2+64=x^2-(8i)^2=(x-8i)(x+8i)$

$16x^2+49=(4x)^2-(7i)^2=(4x-7i)(4x+7i)$

$(x+9i)^2=x^2+2x\cdot9i+(9i)^2=x^2+18xi+81i^2=x^2+18xi-81=x^2-81+18xi$

$(x-2i)^2=x^2-2x\cdot2i+(2i)^2=x^2-4xi+4i^2=x^2-4xi-4=x^2-4-4xi$

$(x+(3+5i))^2=x^2+2x(3+5i)+(3+5i)^2\\\\=x^2+6x+10xi+3^2+2\cdot3\cdot5i+(5i)^2\\\\=x^2+6x+10xi+9+30i+25i^2=x^2+6x+10xi+9+30i-25\\\\=x^2+6x+10xi+30i-16=x^2+6x-16+(10x+30)i$


$Used:\\i^2=-1\\\\a^2-b^2=(a-b)(a+b)\\\\(a+b)^2=a^2+2ab+b^2$