Respuesta :

ItzBlueish

Answer:

[tex]\left(y-8\right)[/tex]

Step-by-step explanation:

[tex]y^2-10y+16[/tex]

[tex]=\left(y^2-2y\right)+\left(-8y+16\right)[/tex]

[tex]y^2-2y[/tex]

[tex]=yy-2y[/tex]

[tex]=y\left(y-2\right)[/tex]

[tex]-8y+16[/tex]

[tex]=-8y+8\cdot \:2[/tex]

[tex]=-8\left(y-2\right)[/tex]

[tex]=y\left(y-2\right)-8\left(y-2\right)[/tex]

[tex]=\left(y-2\right)\left(y-8\right)[/tex]

The other factor is:

[tex]\left(y-8\right)[/tex]

larawooller1
Incase it helps to have a slightly different approach, the way we learnt it is:

y2 - 10y + 16 = 0

When these are factorised, the brackets you have will need to expand to equal the above.

You need to find 2 numbers that multiply together to make the end number, c, and those 2 numbers must also add together to made the middle number, b.

So for this case, those two numbers would be -2 and -8.

Then these go into the brackets,
(y - 2) (y - 8) = 0

This can then be proven to be correct via expansion,

y x y = y2
y x -8 = -8y
-2 x y = -2y
-2 x -8 = 16

(y2) + (-8y) + (-2y) + (16) = 0

y2 - 8y - 2y + 16 = 0

y2 - 10y + 16 = 0

Hope this helps a little?

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