Complex Numbers – Product of Linear Factors 1

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Question 1 of 4

1. Question
Write out as a product of linear factors.

$x^{2} + 16$ $x^2+16$
- 1. $(x + 16 i) (x + 16 i)$ $(x+16i)(x+16i)$
- 2. $(x - 4 i) (x + 4 i)$ $(x-4i)(x+4i)$
- 3. $(x - 16 i) (x + 16 i)$ $(x-16i)(x+16i)$
- 4. $(x + 4 i) (x + 4 i)$ $(x+4i)(x+4i)$
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The products of linear factors can be found using the formulas $a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$ and $a^{2} + b^{2} = (a - b i) (a + b i)$ $a^2+b^2=(a-bi)(a+bi)$ .

To write out $x^{2} + 16$ $x^2+16$ as a product of linear factors, use the formula $a^{2} + b^{2} = (a - b i) (a + b i)$ $a^2+b^2=(a-bi)(a+bi)$ where $a = \sqrt{x^{2}} = x$ $a=sqrt(x^2)=x$ and $b = \sqrt{16} = 4$ $b=sqrt(16)=4$ .

$x^{2} + 16$ $x^2+16$

$=$ $=$ $(x - 4 i) (x + 4 i)$ $(x-4i)(x+4i)$

$(x - 4 i) (x + 4 i)$ $(x-4i)(x+4i)$
Question 2 of 4

2. Question
Write out as a product of linear factors.

$x^{2} + 3$ $x^2+3$
- 1. $(x - i \sqrt{3}) (x + i \sqrt{3})$ $(x-isqrt3)(x+isqrt3)$
- 2. $(2 x + i \sqrt{3}) (2 x - i \sqrt{3})$ $(2x+isqrt3)(2x-isqrt3)$
- 3. $(x - i \sqrt{3}) (x - i \sqrt{3})$ $(x-isqrt3)(x-isqrt3)$
- 4. $(x + i \sqrt{3}) (x + i \sqrt{3})$ $(x+isqrt3)(x+isqrt3)$
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The products of linear factors can be found using the formulas $a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$ and $a^{2} + b^{2} = (a - b i) (a + b i)$ $a^2+b^2=(a-bi)(a+bi)$ .

To write out $x^{2} + 3$ $x^2+3$ as a product of linear factors, use the formula $a^{2} + b^{2} = (a - b i) (a + b i)$ $a^2+b^2=(a-bi)(a+bi)$ where $a = \sqrt{x^{2}} = x$ $a=sqrt(x^2)=x$ and $b = \sqrt{3}$ $b=sqrt3$ .

$x^{2} + 3$ $x^2+3$

$=$ $=$ $(x - i \sqrt{3}) (x + i \sqrt{3})$ $(x-isqrt3)(x+isqrt3)$

$(x - i \sqrt{3}) (x + i \sqrt{3})$ $(x-isqrt3)(x+isqrt3)$
Question 3 of 4

3. Question
Write out as a product of linear factors.

$4 x^{2} + 1$ $4x^2+1$
- 1. $(2 x + i) (2 x + i)$ $(2x+i)(2x+i)$
- 2. $(2 x - i) (2 x + i)$ $(2x-i)(2x+i)$
- 3. $(4 x - i) (4 x + i)$ $(4x-i)(4x+i)$
- 4. $(2 x - i) (2 x - i)$ $(2x-i)(2x-i)$
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The products of linear factors can be found using the formulas $a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$ and $a^{2} + b^{2} = (a - b i) (a + b i)$ $a^2+b^2=(a-bi)(a+bi)$ .

To write out $4 x^{2} + 1$ $4x^2+1$ as a product of linear factors, use the formula $a^{2} + b^{2} = (a - b i) (a + b i)$ $a^2+b^2=(a-bi)(a+bi)$ where $a = \sqrt{4 x^{2}} = 2 x$ $a=sqrt(4x^2)=2x$ and $b = \sqrt{1} = 1$ $b=sqrt(1)=1$ .

$4 x^{2} + 1$ $4x^2+1$

$=$ $=$ $(2 x - i) (2 x + i)$ $(2x-i)(2x+i)$

$(2 x - i) (2 x + i)$ $(2x-i)(2x+i)$
Question 4 of 4

4. Question
Write out as a product of linear factors.

$5 x^{2} + 9$ $5x^2+9$
- 1. $(\sqrt{5} x - 3) (\sqrt{5} x + 3)$ $(sqrt5x-3)(sqrt5x+3)$
- 2. $(\sqrt{5} x + 3) (\sqrt{5} x + 3)$ $(sqrt5x+3)(sqrt5x+3)$
- 3. $(\sqrt{5 x} - 3) (\sqrt{5 x} - 3)$ $(sqrt(5x)-3)(sqrt(5x)-3)$ `
- 4. $(\sqrt{5 x} - 3) (\sqrt{5 x} + 3)$ $(sqrt(5x)-3)(sqrt(5x)+3)$
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The products of linear factors can be found using the formulas $a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$ and $a^{2} + b^{2} = (a - b i) (a + b i)$ $a^2+b^2=(a-bi)(a+bi)$ .

To write out $5 x^{2} + 9$ $5x^2+9$ as a product of linear factors, use the formula $a^{2} + b^{2} = (a - b i) (a + b i)$ $a^2+b^2=(a-bi)(a+bi)$ where $a = \sqrt{5 x^{2}} = \sqrt{5} x$ $a=sqrt(5x^2)=sqrt5x$ and $b = \sqrt{9} = 3$ $b=sqrt(9)=3$ .

$5 x^{2} + 9$ $5x^2+9$

$=$ $=$ $(\sqrt{5} x - 3) (\sqrt{5} x + 3)$ $(sqrt5x-3)(sqrt5x+3)$

$(\sqrt{5} x - 3) (\sqrt{5} x + 3)$ $(sqrt5x-3)(sqrt5x+3)$

Complex Numbers – Product of Linear Factors 1

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1. Question

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2. Question

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3. Question

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4. Question

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Quizzes

		$x^{2} + 16$ $x^2+16$
	$=$ $=$	$(x - 4 i) (x + 4 i)$ $(x-4i)(x+4i)$

		$x^{2} + 3$ $x^2+3$
	$=$ $=$	$(x - i \sqrt{3}) (x + i \sqrt{3})$ $(x-isqrt3)(x+isqrt3)$

		$4 x^{2} + 1$ $4x^2+1$
	$=$ $=$	$(2 x - i) (2 x + i)$ $(2x-i)(2x+i)$

		$5 x^{2} + 9$ $5x^2+9$
	$=$ $=$	$(\sqrt{5} x - 3) (\sqrt{5} x + 3)$ $(sqrt5x-3)(sqrt5x+3)$