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3 years ago

Factor Polynomial 3y^2+y-14

Mathematics

1 answer:

iris [78.8K]3 years ago

4 0

Answer:

Factored Form: ( y - 2 )( 3y + 7 )

Step-by-step explanation:

1. Let us first write down the problem at hand: 3y^2 + y - 14

2. Now let us break this expression into groups:

3y^2 - 6y + 7y - 14 ⇒ ( 3y^2 - 6y )( 7y - 14 )

3. Factor 3y from 3y^2 - 6y:

3y^2 - 6y ⇒ 3y( y - 2 )

4. Factor 7 from 7y - 14:

7y - 14 ⇒ 7( y - 2 )

5. Substitute Step #3, 4 ⇒ Step #2:

3y( y - 2 ) + 7( y - 2 )

6. Factor common term y - 2:

Answer: ( y - 2 )( 3y + 7 )

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Answer:

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Step-by-step explanation:

A parabola is written in the form

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where:

$h$ is the x-coordinate of the vertex of the parabola

$ak$ is the y-coordinate of the vertex of the parabola

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For the parabola in the problem, we know that the vertex has coordinates (4,-3), so we have:

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From this last equation, we get that $a=\frac{-3}{k}$ (3)

Substituting (2) and (3) into (1) we get the new expression:

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We also know that the parabola contains the point (2,-1), so we can substitute

x = 2

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Into eq.(4) and find the value of k:

$-1=-\frac{3}{k}(2-4)^2-3\\-1=-\frac{3}{k}\cdot 4 -3\\2=-\frac{12}{k}\\k=-\frac{12}{2}=-6$

So we also get:

$a=-\frac{3}{k}=-\frac{3}{-6}=\frac{1}{2}$

So the equation of the parabola is:

$f(x)=\frac{1}{2}((x-4)^2 -6)$ (5)

Now we want to rewrite it in the standard form, i.e. in the form

$f(x)=ax^2+bx+c$

To do that, we simply rewrite (5) expliciting the various terms, we find:

$f(x)=\frac{1}{2}((x^2-8x+16)-6)=\frac{1}{2}(x^2-8x+10)=\frac{1}{2}x^2-4x+5$

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Step-by-step explanation:

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First let's simplify the equation so it will be easier to PEMDAS the equation later.

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