Question

please solve this problem as early as possible. ​

Answer 1

AnsWer :

1/√3 , -1/√3 , 2 and 3.

Given :

• p(x) 3x⁴-15x³+17x²+5x-6.
• It have two zeros, 1/√3 and -1/√3.

Solution :

When,

$\tt p(x) = 0.$

$\tt \dashrightarrow x = \pm \frac{1}{ \sqrt{3} }$

There, product

$\tt \dashrightarrow (x + \frac{1}{ \sqrt{3} })(x - \frac{1}{ \sqrt{3} } ) = 0.$

$\tt \dashrightarrow ( {x}^{2} + \frac{x}{ \sqrt{3} } - \frac{x}{ \sqrt{3} } - \frac{1}{3} ) = 0.$

$\tt \dashrightarrow {x}^{2} - \frac{1}{3} = 0.$

$\tt \dashrightarrow 3 {x}^{2} - 1 = 0.$

$\rule{200}3$

Now,p(x) Divided by 3x²-1. We get

* All process provide in attachment

$\tt \dashrightarrow Q= {x}^{2} - 5x + 6.$

$\tt \dashrightarrow R= 0.$

Now, Taking Q.

$\tt \dashrightarrow {x}^{2} - 5x + 6 = 0.$

$\tt \dashrightarrow {x}^{2} - 3x - 2x + 6 = 0.$

$\tt \dashrightarrow x(x - 3) - 2(x - 3) = 0.$

$\tt \dashrightarrow (x - 2)(x - 3) = 0.$

$\begin{array}{r | l} 2 & 6 \\ \cline{2-2} 3 & 3 \\ \cline{2-2} & 1 \end{array}$

Either,

$\tt \dashrightarrow x - 2= 0.$

$\tt \dashrightarrow x = 2.$

And.

$\tt \dashrightarrow x - 3 = 0.$

$\tt \dashrightarrow x = 3.$

Therefore, All the zeros of p(x) is √3 , -√3 , 2 and 3.

Answer 2

factors of the given p(x),(x+1/√3)(x-1/√3)=(3x^2-1)

dividing p(x) with this

we get quotient as

x^2-5x+6

now

P(x)=(x^2-5x+6)(3x^2-1)

p(x)=(x-2)(x-3)(x-1/√3)(x+1/√3)

equating to zero

we will get zeroes as

2,3,1/√3,-1/√3