Question

Obtain all zeros of 2x^4+x³-6x²-3x, if two of its zeros are √3 and -√3.

Answer 1

Answer:

Step-by-step explanation:

As there are 2 zeroes √3 and -√3

So( x-√3) and (x+√3) are factors of p(x)

(a-b)(a+b)=a^2-b^2

So (x+√3)( x-√3)=x^2-3

Divide 2x^4+x³-6x²-3x by x^2-3 to obtain all zeroes

When we divide we get answer as 2x^2+x(quotient)

Zeroes of 2x^2+x are 0,-1/2

Hence 0,-1/2 are other zeroes of p(x)

Answer 2

Answer:

0,-1/2

Step-by-step explanation:

• As there are 2 zeroes √3 and -√3
• So( x-√3) and (x+√3) are factors of p(x)
• (a-b)(a+b)=a^2-b^2
• So (x+√3)( x-√3)=x^2-3
• Divide 2x^4+x³-6x²-3x by x^2-3 to obtain all zeroes
• When we divide we get answer as 2x^2+x(quotient)
• Zeroes of 2x^2+x are 0,-1/2
• Hence 0,-1/2 are other zeroes of p(x)