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2 years ago

11

Factor 2x^4 + 2x^3 - x^2 - x

1 answer:

MAVERICK [17]2 years ago

8 0

2x^3(x+1)-x(x+1)
(2x^3-x)(x+1)
x(2x^2-1)(x+1)

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If f(x) = 2x² – 2. what is f(-6)?<br> If<br> -78<br> 30<br> 18<br> 0 78<br><br> PLS HELP

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Answer:

70

Step-by-step explanation:

Substitute x with -6:

2(-6)²-2

2(36)-2

72-2

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Makovka662 [10]

Let $u=x^2-4$ and $v=4x-5$ . By the product rule,

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=\dfrac{\mathrm d(u^5)}{\mathrm dx}v^4+u^5\dfrac{\mathrm d(v^4)}{\mathrm dx}$

By the power rule, we have $(u^5)'=5u^4$ and $(v^4)'=4v^3$ , but $u,v$ are functions of $x$ , so we also need to apply the chain rule:

$\dfrac{\mathrm d(u^5)}{\mathrm dx}=5u^4\dfrac{\mathrm du}{\mathrm dx}$

$\dfrac{\mathrm d(v^4)}{\mathrm dx}=4v^3\dfrac{\mathrm dv}{\mathrm dx}$

and we have

$\dfrac{\mathrm du}{\mathrm dx}=2x$

$\dfrac{\mathrm dv}{\mathrm dx}=4$

So we end up with

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=10xu^4v^4+16u^5v^3$

Replace $u,v$ to get everything in terms of $x$ :

$\dfrac{\mathrm d((x^2-4)^5(4x-5)^4)}{\mathrm dx}=10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3$

We can simplify this by factoring:

$10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3=2(x^2-4)^4(4x-5)^3\bigg(5x(4x-5)+8(x^2-4)\bigg)$

$=2(x^2-4)^4(4x-5)^3(28x^2-57)$

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2 years ago

Pls only do this if yk it because i’m giving correct answer brainliest! :))

Leno4ka [110]

Answer:

$x=2+\frac{1}{2}\sqrt[]{21}$

or

$x=2-\frac{1}{2}\sqrt{{21}$

Step-by-step explanation:

$4x^2-16x-26=-21$

Add 21 on both sides.

$4x^2-16x-26+21=-21+21$

$4x^2-16x-5=0$

a=4

b=-16

c=-5

$x=\frac{-b\frac{+}{}\sqrt[]{b^2-4ac} }{2a}$

$x=\frac{-(-16)\frac{+}{}\sqrt[]{(-16)^2-4(4)(-5)} }{2(4)}$

$x=\frac{16\frac{+}{}\sqrt[]{256+80} }{8}$

$x=\frac{16\frac{+}{}\sqrt[]{336} }{8}$

$x=\frac{16\frac{+}{}\sqrt[]{2^2*2^2*21} }{8}$

$x=\frac{16\frac{+}{}2*2\sqrt[]{21} }{8}$

$x=\frac{16\frac{+}{}4\sqrt[]{21} }{8}$

---------------------------------------------------------------------------

$x=\frac{16}{8}+\frac{4\sqrt[]{21}}{8}$

$x=2+\frac{1}{2}\sqrt[]{21}$

---------------------------------------------------------------------------

$x=\frac{16}{8}-\frac{4\sqrt[]{21}}{8}\\x=2-\frac{1}{2}\sqrt{{21}$

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