Thread: How to factorize 4z^2 -4z+17 Over C?

1. How to factorize 4z^2 -4z+17 Over C?

By over C I mean it has a solution that is a complex number

My Attempt:

4z^2-4z+17=(4z^2-4z+4)-4+17
................=(2z-2)^2 +13
................=(2z-2)^2 - (Squareroot of (13)*i)^2
................=((2z+2)+(Squareroot of 13 *i))((2z+2)-(Squareroot of 13 *i))

However, the real answer is (2z-1+4i)(2z-1-4i)

I can factorise questions such as z^2-4z+17 but when there is a number infront of z^2 I do not know how to do it.

Thank you for any help .

2. If $z_1,z_2$ are the solutions of :

$4z^2-4z+17=0$

then,

$4z^2-4z+17=4(z-z_1)(z-z_2)$

Fernando Revilla

3. Originally Posted by anees7112
By over C I mean it has a solution that is a complex number

My Attempt:

4z^2-4z+17=(4z^2-4z+4)-4+17
................=(2z+2)^2 +13
................=(2z+2)^2 - (Squareroot of (13)*i)^2
................=((2z+2)+(Squareroot of 13 *i))((2z+2)-(Squareroot of 13 *i))

However, the real answer is (2z-1+4i)(2z-1-4i)

I can factorise questions such as z^2-4z+17 but when there is a number infront of z^2 I do not know how to do it.

Thank you for any help .
Alternatively

$4z^2-4z+17=4z^2-4z+1+16=(2z-1)(2z-1)+16=(2z-1)^2-(4i)^2$

which is the difference of two squares

4. Quadratic Formula works nicely here

$\displaystyle z = \frac{4 \pm \sqrt{(-4)^2 - 4\cdot 4\cdot 17}}{2\cdot 4}$.

Simplify

5. Originally Posted by Archie Meade
Alternatively

$4z^2-4z+17=4z^2-4z+1+16=(2z-1)(2z-1)+16=(2z-1)^2-(4i)^2$

which is the difference of two squares
Hello.
I am just wondering how you got 4z^2-4z+1+16. My teacher told me to take the -4 from -4z , half it, and then square it. (-4)/2=-2 2^2=4.
Thank you.

6. Originally Posted by anees7112
Hello.
I am just wondering how you got 4z^2-4z+1+16. My teacher told me to take the -4 from -4z , half it, and then square it. (-4)/2=-2 2^2=4.
Thank you.
That method applies well when the "squared term" has a coefficient of 1.

For example

$z^2-4z+17=z^2-4z+4+13=(z-2)^2+13$

However, because of the $4z^2$ term

$(2z-1)(2z-1)=2z(2z-1)-1(2z-1)=4z^2-2z-2z+1=4z^2-4z+1$

so I took 1 from 17 instead.

7. Originally Posted by Archie Meade
That method applies well when the "squared term"

so I took 1 from 17 instead.
Alright thanks for that I think I understand now.

But one more thing. How did you chose the numbers 1 and 17 exactly. I know they work but how did you pick these 2.

I tried your method with 9z-24z+32
and came up with 9z-24z+16 +16
But it took me a while to come up with 16 and 16.

Is there a formula like the (x*0.5)^2 for the case where the coefficient of z^2 is 1, to work out 1 and 17 or 16 and 16.

Thanks

8. Originally Posted by anees7112
Alright thanks for that I think I understand now.

But one more thing. How did you chose the numbers 1 and 17 exactly. I know they work but how did you pick these 2.

I tried your method with 9z^2-24z+32
and came up with 9z^2-24z+16 +16
But it took me a while to come up with 16 and 16.

Is there a formula like the (x*0.5)^2 for the case where the coefficient of z^2 is 1, to work out 1 and 17 or 16 and 16.

Thanks
Ok,
I will lead you through both of those.

(1)

$4z^2-4z+17$

will factor with $(4z+?)(z+?)\;\;\;or\;\;\;(2z+?)(2z+?)$

I must end up with $-4z$ as my centre term after multiplying out the factors,
so that rules out the first pair if I'm looking for integers.
Hence

$(2z+?)(2z+?)$ must give me $4z^2-4z$

which is $4z^2-2z-2z$ so now I see what my $"?"$ must be. I need $-1$ there.

Then I just break up 17 into 16+1, since $(-1)^2=1$.

Another way to look at would have been to figure that 16 is the nearest square to 17
and so is a good place to start.

(2)

$9z^2-24z+32$

The most obvious place to start is

$(3z)^2=9z^2$

and

$-24z=-12z-12z$

$3(4)=12\Rightarrow\ (3z-4)(3z-4)=3z(3z-4)-4(3z-4)=9z^2-12z-12z+16$

Hence we take 16 from 32 to get

$9z^2-24z+32=9z^2-24z+16+16=(3z-4)^2-(4i)^2=(3z-4+4i)(3z-4-4i)$