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Len [333]
3 years ago
10

I need help with this Factoring Quadratics question.Can someone help me?

Mathematics
1 answer:
antoniya [11.8K]3 years ago
7 0

If you mean "factor over the rational numbers", then this cannot be factored.

Here's why:

The given expression is in the form ax^2+bx+c. We have

a = 3

b = 19

c = 15

Computing the discriminant gives us

d = b^2 - 4ac

d = 19^2 - 4*3*15

d = 181

Note how this discriminant d value is not a perfect square

This directly leads to the original expression not factorable

We can say that the quadratic is prime

If you were to use the quadratic formula, then you should find that the equation 3x^2+19x+15 = 0 leads to two different roots such that each root is not a rational number. This is another path to show that the original quadratic cannot be factored over the rational numbers.

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Determine if "h=3" is a solution.<br> 3h=2(h+2)-1
lutik1710 [3]

Answer:

Yes is a solution

Step-by-step explanation:

3h = 2(h+2)-1

We replace h=3

3*3 = 2(3+2) - 1

Solve

9 = 2(5)-1

9 = 10 - 1

9 = 9

Is correct h= 3

when we manage to find the same value on both sides of equality it means that the value entered is a possible option

4 0
3 years ago
Read 2 more answers
Given that f(x) = x2 + 2x + 3 and g(x) = quantity of x plus four, over three , solve for f(g(x)) when x = 2.
Sidana [21]
f(x)=x^2+2x+3
g(x)=\frac{x+4}{3}
f(g(x))=(\frac{x+4}{3})^2+2(\frac{x+4}{3})+3
f(g(x))=\frac{x^2+8x+16}{9}+\frac{2x+8}{3}+3
f(g(2))=\frac{2^2+8(2)+16}{9}+\frac{2(2)+8}{3}+3
f(g(2))+\frac{4+16+16}{9}+\frac{4+8}{3}+3
f(g(2))=\frac{36}{9}+\frac{12}{3}+3
f(g(2))=4+4+3
f(g(2)=11

Check:
g(2)=(2+4)/3=2
f(2)=2^2+2(2)+3=4+4+3=11  Checks. 
4 0
3 years ago
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Write a polynomial function of least degree with integral coefficients that has the
frutty [35]

A polynomial function of least degree with integral coefficients that has the

given zeros  f(x)=x^4+x^3+9x^2+9x

Given

Given zeros are 3i, -1  and 0

complex zeros occurs in pairs. 3i is one of the zero

-3i is the other zero

So zeros are 3i, -3i, 0 and -1

Now we write the zeros in factor form

If 'a' is a zero then (x-a) is a factor

the factor form of given zeros

\:\left(x-3i\right)\left(x-\left(-3i\right)\right)\left(x-0\right)\left(x-\left(-1\right)\right)\\\left(x-3i\right)\left(x+3i\right)\left(x-0\right)\left(x+1\right)

Now we multiply it to get the polynomial

x\left(x-3i\right)\left(x+3i\right)\left(x+1\right)\\x\left(x^2+9\right)\left(x+1\right)\\x\left(x^3+x^2+9x+9\right)\\x^4+x^3+9x^2+9x

polynomial function of least degree with integral coefficients that has the

given zeros  f(x)=x^4+x^3+9x^2+9x

Learn more : brainly.com/question/7619478

6 0
3 years ago
A company that manufactures cell phones has been given a quarterly operating budget of $1,176,912.42. The company's quarterly op
amm1812

Answer:

maximum is 16,853 cell phones

≤  18.853

Step-by-step explanation:

1,176,912.42.   per quarter budget

247,638.00    per quarter fixed cost

We subtract then divide the amount cost of phones and ensure its less or rounded down.

1,176,912.42 -247,638 = 929274.42

929274.42 / 55.14 = 16853

16,852 < m ≤ 16,853

6 0
2 years ago
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First row is 3 cookies for every $1, proportional, and Table to the right
Second row is 3 more cookies than donuts, non-proportional, and the Table on the left
5 0
3 years ago
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