Select the two binomials that are factors of this trinomial x^20 - x - 20

Mathematics

1 answer:

Mashcka [7]3 years ago

8 0

Answer:

$(x-5)(x+4)$

Step-by-step explanation:

To factor the equation, break it into two binomials which multiply to make the equation. To write these binomials (x+a)(x+b), find factors which multiply to -20 and add to -1 for a and b.

20: 1, 2, 4, 5, 10, 20

-5+4 = -1

$x^2-x-20 = (x-5)(x+4)$

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Verify cot x sec^4x=cotx +2tanx +tan^3x

Tanzania [10]

Answer:

See explanation

Step-by-step explanation:

We want to verify that:

$\cot(x) \: { \sec}^{4} x = \cot(x) + 2 \tan(x) + { \tan}^{3} x$

Verifying from left, we have

$\cot(x) \: { \sec}^{4} x = \cot(x) \: ( 1 + { \tan}^{2} x )^{2}$

Expand the perfect square in the right:

$\cot(x) \: { \sec}^{4} x = \cot(x) \: ( 1 + { 2\tan}^{2} x + { \tan}^{4} x)$

We expand to get:

$\cot(x) \: { \sec}^{4} x = \cot(x) \: + \cot(x){ 2\tan}^{2} x +\cot(x) { \tan}^{4} x$

We simplify to get:

$\cot(x) \: { \sec}^{4} x = \cot(x) \: + 2 \frac{ \cos(x) }{\sin(x) ) } \times \frac{{ \sin}^{2} x}{{ \cos}^{2} x} +\frac{ \cos(x) }{\sin(x) ) } \times \frac{{ \sin}^{4} x}{{ \cos}^{4} x}$