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2 years ago

How do I factor 14x^2-7xy-147y^2

Mathematics

1 answer:

Viefleur [7K]2 years ago

6 0

STEP 1

Equation at the end of step

((14 • (x2)) - 7xy) - (3•72y2)

Equation at the end of step 2

((2•7x2) - 7xy) - (3•72y2)

STEP 3

STEP 4

Pulling out like terms

4.1 Pull out like factors :

14x2 - 7xy - 147y2 = 7 • (2x2 - xy - 21y2)

Trying to factor a multi variable polynomial :

4.2 Factoring 2x2 - xy - 21y2

Try to factor this multi-variable trinomial using trial and error

Found a factorization : (2x - 7y)•(x + 3y)

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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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ight

Step-by-step explanation:

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