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3 years ago

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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Answer:

Yes is a solution

Step-by-step explanation:

3h = 2(h+2)-1

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

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

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

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

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