Question

Complete the following exercises by applying polynomial identities to complex numbers.

Answer 1

Answer:

(a)

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

(b)

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

(c)

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

(d)

$(x-2i)^2=x^2-4xi-4$

(e)

$(x+(3+5i))^2=10ix+30i+25i^2+x^2+6x+9$

Step-by-step explanation:

(a)

we are given

$x^2+64$

we can also write as

$(x)^2+8^2$

$(x)^2-(8i)^2$

now, we can use factor formula

$a^2-b^2=(a-b)(a+b)$

we get

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

(b)

we are given

$16x^2+49$

we can also write as

$(4x)^2+7^2$

$(4x)^2-(7i)^2$

now, we can use factor formula

$a^2-b^2=(a-b)(a+b)$

we get

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

(c)

$(x+9i)^2$

we can use formula

$(a+b)^2=(a^2+2ab+b^2)$

so, we can write as

$(x+9i)^2=x^2+2x\cdot \:9i+\left(9i\right)^2$

we can simplify it

and we get

$=x^2+18ix-81$

$=x^2+18xi-81$

(d)

$(x-2i)^2$

we can use formula

$(a+b)^2=(a^2+2ab+b^2)$

so, we can write as

$(x-2i)^2=x^2-2x\cdot \:2i+\left(2i\right)^2$

we can simplify it

and we get

$=x^2-4ix-4$

$=x^2-4xi-4$

(e)

$(x+(3+5i))^2$

we can distribute

so, we can write as

$=\left(5i+x+3\right)\left(5i+x+3\right)$

$=xx+x\cdot \:3+x\cdot \:5i+3x+3\cdot \:3+3\cdot \:5i+5ix+5i\cdot \:3+5i\cdot \:5i$

$=xx+3x+5ix+3x+3\cdot \:3+3\cdot \:5i+5ix+5\cdot \:3i+5\cdot \:5ii$

now, we can simplify it

$=10ix+30i+25i^2+x^2+6x+9$