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4 years ago

Can some one plz answer this question, factories y^2-8y+12

Mathematics

1 answer:

Blababa [14]4 years ago

6 0

Answer:

(y - 2)(y - 6)

Step-by-step explanation:

To factorise

Consider the factors of the constant term (+ 12) which sum to give the coefficient of the y- term (- 8)

The factors are - 2 and - 6

since - 2 × - 6 = + 12 and - 2 - 6 = - 8, hence

y² - 8y + 12 = (y - 2)(y - 6)

You might be interested in

Lin and Priya were working on solving this system of equations.

garri49 [273]

Answer:

we can conclude two things that:

If we multiple the two sides of any given equation by the same factor, we would get an equivalent equation, which will have the same solution as the original solution.
Either person's move will work. Lin's move eliminated the x variable, while Priya's eliminated y variable, but in the end the solution was same.

Step-by-step explanation:

Why either move creates a new equation with the same solutions as the original equation?

If we multiple the two sides of any given equation by the same factor, we would get an equivalent equation, which will have the same solution as the original solution.

When we multiple the two sides of any given equation by the same number, it would keep the two sides of that particular equation equal. So, whatever the the solution the first equation may get, will still work for the second equation.

Determining Lin's first move i.e. to multiply the first equation by 3.

Let us consider the equation

x/3 + 2y = 4 .....[1]

x + y = -3 .....[2]

Lin's first move is to multiply the first equation by 3.

3(x/3 + 2y) = 3(4 )

x + 6y = 12 .....[3]

Now subtract the Equation [2] from Equation [3]

x + 6y - x - y = 12 - (-3)

5y = 15

y = 3

Putting y = 3 in [2]

x + (3) = -3

x = -6

So, x = -6 and y = 3

Determining Priya's first move i.e. to multiply the Second equation by 2.

Let us consider the equation

x/3 + 2y = 4 .....[1]

x + y = -3 .....[2]

Priya's first move is to multiply the second equation by 2.

2(x + y) = 2(-3)

2x + 2y = -6 .....[3]

Now subtract the Equation [2] from [1]

x/3 + 2y - 2x - 2y= 4 - (-6)

x/3 - 2x = 10

x - 6x = 30

x = -6

Putting x = -6 in Equation [2]

x + y = -3

-6 + y = -3

y = 3

So, x = -6 and y = 3

So, from the entire analysis, we can conclude two things that:

If we multiple the two sides of any given equation by the same factor, we would get an equivalent equation, which will have the same solution as the original solution.
Either person's move will work. Lin's move eliminated the x variable, while Priya's eliminated y variable, but in the end the solution was same.

Keywords: system of equation, solution, equation

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4 0

3 years ago

Which equation are true for x=-2 and x=2 check all that apply

ki77a [65]

1 it is true
${x}^{2} - 4 = 0 \\ {x}^{2} = + 4 \\ x = \sqrt{4} = 2$
2 it's not true
${x}^{2} = - 4$
3 it's not true
$3 {x}^{2} + 12 = 0 \\ 3 {x}^{2} = - 12 \\ {x}^{2} = \frac{ - 12}{3} = - 4 \\$
4 it's true
$4 {x}^{2} = 16 \\ {x}^{2} = \frac{16}{4} = 4 \\ x = \sqrt{4} = 2$
5- it's true
${2(x - 2)}^{2} = 0 \\ {(x - 2)}^{2} = 0 \\ x - 2 = 0 \\ x = 2$