Question

For the polynomial p(y) = y3 – 2y2 – 297 - 42, one
of its factors is​

Answer 1

Step-by-step explanation:

By factorization of f(-2)=(-2)^{3}-2(-2)^{2}-29(-2)-42f(−2)=(−2)

3

−2(−2)

2

−29(−2)−42 , the factors are (y+3)(y-7)(y+2)

Solution:

By trial, f(1) ≠ 0,f(-1) ≠ 0,f(2) ≠ 0,f(-2) = 0

Hence y+2 is a factor.

f(-2)=(-2)^{3}-2(-2)^{2}-29(-2)-42f(−2)=(−2)

3

−2(−2)

2

−29(−2)−42

f(-2) = (-8) – 2(4) + 58 – 42

f(-2) = -16 + 58 – 42

f(-2) = 0

The long division process is shown in the below attached image.

\left(y^{2}-4 y-21\right)(y+2)(y

2

−4y−21)(y+2) are the factors.

=\left(y^{2}-7 y+3 y-21\right)(y+2)=(y

2

−7y+3y−21)(y+2)

= y(y – 7) + 3(y – 7)( y+2)

(y+3)(y-7)(y+2) are the factors.