Question

Factorise the following polynomials 28c2d2 – 21c2d​

Answer 1

Step-by-step explanation:

Let the given polynomial be p(x)=2x

3

−9x

2

+7x+6.

We will now substitute various values of x until we get p(x)=0 as follows:

Forx=0

p(0)=2(0)

3

−9(0)

2

+(7×0)+6=0−0+0+6=6



=0

∴p(0)



=0

Forx=1

p(1)=2(1)

3

−9(1)

2

+(7×1)+6=2−9+7+6=15−9=6



=0

∴p(1)



=0

Forx=2

p(2)=2(2)

3

−9(2)

2

+(7×2)+6=(2×8)−(9×4)+14+6=16−36+14+6=36−36=0

∴p(2)=0

Thus, (x−2) is a factor of p(x).

Now,

p(x)=(x−2)⋅g(x).....(1)

⇒g(x)=

(x−2)

p(x)

Therefore, g(x) is obtained by after dividing p(x) by (x−2) as shown in the above image:

From the division, we get the quotient g(x)=2x

2

−5x−3 and now we factorize it as follows:

2x

2

−5x−3

=2x

2

−6x+x−3

=2x(x−3)+1(x−3)

=(x−3)(2x+1)

From equation 1, we get p(x)=(x−2)(x−3)(2x+1).

Hence, 2x

3

−9x

2

+7x+6=(x−2)(x−3)(2x+1).