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expeople1 [14]
3 years ago
14

Please help 5x^2+21x+4

Mathematics
2 answers:
SVEN [57.7K]3 years ago
8 0

Answer:

(5x+1) (x+4)

Step-by-step explanation:

Daniel [21]3 years ago
5 0
I think it’s (x+4)*(5x+1)

Write 21x as a sum ~ 20x + x

5x^2+20x+x+4

Then you factor out the 5x from the expression

5x * (x+4) + x+4

Then you factor out x+4 from the expression

(X+4) * (5x+1)

Then that’s your answer :) have a great day
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Answer:

The general form of a quadratic function is --> y = ax2 + bx + c

Since (0, -2) exists for the function, we can plug in those values:

-2 = a(0)2 + b(0) + c

-2 = 0 + 0 + c

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So now our function so far is --> y = ax2 + bx - 2

We have two pairs of coordinates left: (-1, -8) and (3, -8).

First, plug in the first pair and simplify as much as you can:

-8 = a(-1)2 + b(-1) - 2

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Second, plug in the second pair and simplify as much as you can:

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Now we have these two equations left:

a - b = -6

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Now we solve for a and b using systems of equations, using one of three ways:

substitution

elimination

graphing (not my favorite, but it is doable)

Using substitution:

a - b = -6 can be rewritten as a = b - 6

plug into the second equation and solve for b

9(b - 6) + 3b = -6 (distribute the 9)

9b - 54 + 3b = -6 (combine all of the b's)

12b - 54 = -6 (add 54 to both sides)

12b = 48 (divide by 12 on both sides to isolate b)

b = 4

plug b into one of the original two equations

a - 4 = -6 (add 4 to both sides)

a = -2

The quadratic equation for this table is y = -2x2 + 4b - 2

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