Question

Factorise 2x3

– 11x

2 + 17x – 6 using factor

theorem.
will mark brainliest if appropriate answer givem​

Answer 1

Step-by-step explanation:

According to Factor theorem, if (x - a) is a polynomial factor f(x), then f(a) = 0

Let f(x) = x^{3}-6 x^{2}+11 x-6f(x)=x3−6x2+11x−6

Let us check if (x - 1) is the factor of f(x),

Then,

f(1) = 1^{3}-6(1^{2})+11(1)-6=1-6+11-6=0f(1)=13−6(12)+11(1)−6=1−6+11−6=0

Therefore (x-1) is a factor of f(x)

Let us check for the other factors

Hence,

f(x)=(x-1)(x^{2}-5 x+6)f(x)=(x−1)(x2−5x+6)

x^{2}-5 x+6=x^{2}-2 x-3 x+6x2−5x+6=x2−2x−3x+6

=x(x-2)-3(x-2)=x(x−2)−3(x−2)

= (x - 2)(x - 3)=(x−2)(x−3)

f(x) = (x - 1)(x - 2)(x - 3)f(x)=(x−1)(x−2)(x−3)

Therefore, 1, 2, 3 are the factors of f(x)

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Answer 2

Answer:

Zeroes of x

3

−6x

2

+11x−6

By using rational theorem, the roots can be among the factors of

1

6

=6

Let us try x=1

⇒(1)

3

−6(1)

2

+11(1)−6=0

∴(x−1) is a factor of x

3

−6x

2

+11x−6.

Now, using synthetic division method :

So, the quotient =x

2

−5x+6

Now, using common factor theorem,

⇒x

2

−5x+6=x

2

−2x−3x+6

=x(x−2)−3(x−2)

=(x−2)(x−3)

∴ Zeroes of the polynomial =1,2,3

So, factor of the polynomial =(x−1)(x−2)(x−3).

Hence, the answer is (x−1)(x−2)(x−3).

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