What is factor (4x^4+8x^2-3)+(8x^3+x) ?

The solution to factor (4x^4+8x^2-3)+(8x^3+x) is (x+1)(4x^3+4x^2+4x-3)

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factor (4x^4+8x^2-3)+(8x^3+x)

factor

(4 x^{4} + 8 x^{2} - 3) + (8 x^{3} + x)

(x + 1) (4 x^{3} + 4 x^{2} + 4 x - 3)

Solution steps

(4 x^{4} + 8 x^{2} - 3) + (8 x^{3} + x)

Remove parentheses:

(a) = a

= 4 x^{4} + 8 x^{2} - 3 + 8 x^{3} + x

Use the rational root theorem

a_{0} = 3, a_{n} = 4

The dividers of

a_{0} : 1, 3,

The dividers of

a_{n} : 1, 2, 4

Therefore, check the following rational numbers:

\pm \frac{1 , 3}{1 , 2 , 4}

- \frac{1}{1}

is a root of the expression, so factor out

x + 1

= (x + 1) \frac{4 x ^{4} + 8 x ^{3} + 8 x ^{2} + x - 3}{x + 1}

\frac{4 x ^{4} + 8 x ^{3} + 8 x ^{2} + x - 3}{x + 1} = 4 x^{3} + 4 x^{2} + 4 x - 3

= (x + 1) (4 x^{3} + 4 x^{2} + 4 x - 3)

x \cdot (\frac{d}{d x} (x^{2}))

60000 0^{0.12 \cdot t}

1 - 4 x - 1

\frac{x ^{2} - 2 x - 3}{x - 3}

2 x - x^{2}

What is factor (4x^4+8x^2-3)+(8x^3+x) ?
The solution to factor (4x^4+8x^2-3)+(8x^3+x) is (x+1)(4x^3+4x^2+4x-3)