Question

if p(y)=y3-4y2-y+6 then show that p(3) =0 and hence factorise the p(y)​

Answer 1

Answer:

p(y)=y³-4y²+y+6

TO SHOW～>`)～～～

p(3) = 0

3³-4(3)² +3 +6

27-36+3+6

0

hence proved,p(3) is a factor of p(y)

now, by hit and trial, p(-1) is also an factor of p(y)

and by long division method we can conclude that,(y-3) and (y-2) are the factors of p(y)

Answer 2

Given :

p(y) = $y^{3} - 4y^{2} - y + 6$

To find :

Prove p(3) = 0

Factorise p(y)

Solution :

First we have to prove p(3) = 0

Take y = 3

Then p(y) = p(3) = $3^{3} - 4(3^{2}) - 3 +6$

                   p(3) = 27 - 36 - 3 + 6 = -6

Since p(3) = -6, p(3) $\neq$ 0

Therefore (y-3) is not a factor of p(y)

Hence we cannot factorise p(y) using (y-3)