Question

If one of the zeroes of the cubic polynomial x3-7x+6 is 2, then the product of the other two zeroes is

Answer 1

Tip:

As we know that, the any Quadratic polynomial $ax^{2} +bx+c=0$

∴ Product of roots: $\alpha \beta$ = $\frac{c}{a}$

To Find :

We have to find the product of other two Zeros.

Explanation:

Step 1:

As we know that, The Given function is

Let, $p(x)=x^{3} -7x+6$

As we have to find the product of other two zeros

The one of the zero is given i.e. 2

Step 2:

So , here let $(x-2)$ be the factor of $p(x)$

∴ Divide $p(x)$ with $x-2$

∴ we will get $g(x)=x^{2} +2x-3$

∴ On further factorization

we will get, $g(x)=(x-1) (x+3)$

∴ we get $\alpha =1$ and $\beta =-3$

so, $\alpha \beta =-3$

Here, the product of other two zeros are -3

Final Answer :

Therefore, we get required solution of above given function $p(x)=x^{3} -7x+6$ is -3

#SPJ2

Answer 2

Answer:

Step 1:

As we know that, The Given function is

Let, p(x)=x^{3} -7x+6p(x)=x3−7x+6

As we have to find the product of other two zeros

The one of the zero is given i.e. 2

Step 2:

So , here let (x-2)(x−2) be the factor of p(x)p(x)

∴ Divide p(x)p(x) with x-2x−2

∴ we will get g(x)=x^{2} +2x-3g(x)=x2+2x−3

∴ On further factorization

we will get, g(x)=(x-1) (x+3)g(x)=(x−1)(x+3)

∴ we get \alpha =1α=1 and \beta =-3β=−3

so, \alpha \beta =-3αβ=−3

Here, the product of other two zeros are -3

Final Answer :

Therefore, we get required solution of above given function p(x)=x^{3} -7x+6p(x)=x3−7x+6 is -3