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3 years ago

5

Please I need answers Factorize x²-6x+9

1 answer:

miskamm [114]3 years ago

4 0

Answer:

The factors of the given equation are:

$x^2-6x+9=(x-3)(x-3)=(x-3)^2$

Step-by-step explanation:

We have :

$x^2-6x+9$

Using middle term splitting theorem to factorize the expression :

$x^2-3x-3x+9$

$x(x-3)+(-3)(x-3)$

$(x-3)(x-3)$

$x^2-6x+9=(x-3)(x-3)=(x-3)^2$

The factors of the given equation are:

$x^2-6x+9=(x-3)(x-3)=(x-3)^2$

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Answer:

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3 years ago

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Answer:

10.89

Step-by-step explanation:

Circle:

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3 years ago

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neonofarm [45]

Answer:

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How to find imaginary zeros anf real zeros of F(x)=-4x^5-8x^3+12x

Fofino [41]

Answer:

$x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}$

Step-by-step explanation:

We are given the function:

$f(x)=-4x^5-8x^3+12x$

And we want to finds its zeros.

Therefore:

$0=-4x^5-8x^3+12x$

Firstly, we can divide everything by -4:

$0=x^5+2x^3-3x$

Factor out an x:

$0=x(x^4+2x^2-3)$

This is in quadratic form. For simplicity, we can let:

$u=x^2$

Then by substitution:

$0=x(u^2+2u-3)$

Factor:

$0=x(u+3)(u-1)$

Substitute back:

$0=x(x^2+3)(x^2-1)$

By the Zero Product Property:

$x=0\text{ and } x^2+3=0\text{ and } x^2-1=0$

Solving for each case:

$x=0\text{ and } x=\pm\sqrt{-3}\text{ and } x=\pm\sqrt{1}$

Therefore, our real and complex zeros are:

$x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}$

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3 years ago