Question

Oʙᴛᴀɪɴ ᴀʟʟ ᴛʜᴇ ᴢᴇʀᴏᴇs
$3x ^{2} \times 6x^{2} - 2x ^{2} - 10x - 5$
ɪғ ᴛᴡᴏ ᴏғ ɪᴛs ᴢᴇʀᴏ ᴀʀᴇ
$\sqrt \frac{5}{3}$
ᴀɴᴅ
$- \sqrt \frac{5}{3}$
​

Answer 1

$\huge\bold{Correct}$

$\huge\bold{\: \: \: Question:-}$

• Obtain all other zeroes of $3 {x}^{4} + 6 {x}^{3} - 2 {x}^{ {2}} - 10x - 5$
• If two zeroes are $\sqrt{ \frac{5}{3} } \: \: and \: \: - \sqrt{ \frac{5}{3} }$

$\huge\bold{Solution:-}$

Let,

$p(x) = 3 {x}^{4} + 6 {x}^{3} - 2 {x}^{2} - 10x - 5$

Given,

$\sqrt{ \frac{5}{3} } \: \: and \: \: - \sqrt{ \frac{5}{3} }$

are roots of p( x )

So,

$(x - \sqrt{{ \frac{5}{3} } } )$

$(x + \sqrt{ \frac{5}{3} })$ are factors of p( x )

Also,

$(x + \sqrt{ \frac{5}{3} } )(x - \sqrt{ \frac{5}{3} } )$is a factor of p( x )

i.e

${x}^{2} - \frac{5}{3}$ is a factor of p ( x )

Divide p( x ) with it's factor

$\frac{3 {x}^{4} + 6 {x}^{3} - 2 {x}^{2} - 10x - 5 }{ {x}^{2} - \frac{5}{3} }$

See the attachment

It gives

$3 {x}^{2} + 6x + 3$

Factorize it

$3 \times 3 = 9$

$3 + 3 = 6$

So,

$3 {x}^{2} + 3x + 3x + 3$

$3x(x + 1) + 3(x + 1)$

$(x + 1)(3x +3 )$

$x + 1 = 0$

$x = - 1$

And

$3x + 3 = 0$

$3x = - 3$

$x = - \frac{ 3}{3} = - 1$

The other factors are -1 and -1

Hope this is helpful✓

Answer 2

Answer:

Refer the above attachment