Using factor theorem, factorize of the polynomials:

Solution:

Given, $f(y)=y^{3}-7 y+6$

The constant term in f(y) is 6

The factors are ± 1, ± 2, ± 3, ± 6

Let, y – 1 = 0

=> y = 1

$f(1)=(1)^{3}-7(1)+6$

= 1 – 7 + 6

= 0

So, (y – 1) is the factor of f(y)

Similarly, (y – 2) and (y + 3) are also the factors

Since, f(y) is a polynomial which has degree 3, it cannot have more than 3 linear factors

=> f(y) = k(y – 1)( y – 2)(y + 3)

$\Rightarrow y^{3}-7 y+6=k(y-1)(y-2)(y+3)--1$

Substitute k = 0 in eq 1

=> 0 – 0 + 6 = k(-1)(-2)(3)

=> 6 = 6k

=> k = 1

$y^{3}-7 y+6=(1)(y-1)(y-2)(y+3)$

$y^{3}-7 y+6=(y-1)(y-2)(y+3)$

Hence, $y^{3}-7 y+6=(y-1)(y-2)(y+3)$