Question

Use factor theorem to find whether polynomial q(x) is a factor of polynomial p(x) or not:

(i) p(x) = x³ - 6x² - 19x - 84 ,q(x) = x - 7
(ii) p(x) = 2x³ - 9x² + x + 12,q(x) = 3 - 2x​

Answer 1

Answer:

Step-by-step explanation:

i)By factor theorem if (x-a) is a factor of a polinomial p(x) then p(a)=0

p(7)=7^3-6(7)^2-19(7)-84

      =343-294-133-84

     =-168≠0

∴ q(X) is not a factor of q(x)

ii)q(x)=3-2x

3-2x=0

-2x=-3

x=3/2

p(3/2)=2(3/2)^3-9(3/2)^2+3/2+12

         =27/4-81/4+3/2+12

         =(27-81+6+48)/4

         =0

∴q(x) is a factor of p(x)

Answer 2

Answer:-

Before checking the given polynomials whether they are factors of the given polynomial, we must know what factor theorem is.

Factor theorem: If p(x) is a polynomial of degree/power greater than or equal to 1 , and a ∈ R (Real number) , then (x - a) is the factor and zero of p(x) if p(a) = 0 (or) p(a) = 0 if (X - a) is a factor of x.

(i) Given :

p(x) = x³ - 6x² - 19x - 84

q(x) = x - 7

Here, x - 7 is a factor of x³ - 6x² - 19x - 84.

Comparing it with x - a we get or taking q(x) = 0 we get,

⟹ a = 7

Now,

We have to prove that,

⟹ p(7) = 0. [ ∵ p(a) = p(7) or p(x) where x = 7. ]

⟹ (7)³ - 6(7)² - 19(7) - 84 = 0

⟹ 343 - 6(49) - 133 - 84 = 0

⟹ 343 - 294 - 133 - 84 = 0

⟹ - 168 ≠ 0

Here, p(7) ≠ 0.

So, x - 7 is not a factor of x³ - 6x² - 19x - 84.

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(ii) p(x) = 2x³ - 9x² + x + 12 ; q(x) = 3 - 2x

Now, taking q(x) = 0 for the value of x we get,

⟹ 3 - 2x = 0

⟹ 3 = 2x

⟹ 3/2 = x

Now,

We have to prove whether,

⟹ p(3/2) = 0

⟹ 2(3/2)³ - 9(3/2)² + 3/2 + 12 = 0

⟹ 2(27/8) - 9(9/4) + 3/2 + 12 = 0

⟹ 27/4 - 81/4 + 3/2 + 12 = 0

⟹ (27 - 81 + 6 + 48) / 4 = 0

⟹ 0/4 = 0

⟹ 0 = 0.

Here, p(3/2) = 0.

So, 3 - 2x is a factor of 2x³ - 9x² + x + 12.