Question

3. find the Value of k , if x - 1 is a factor of p (x) in each of the following cases: i) p(x) = x²+ x+ k ii.) 2x²+ Kx + √2 iii.) 6x² + 5x - 6 iv.) p(x) = kx² - 3x + k​

Answer 1

Answer:

k =

$\frac{3}{2}$

Step-by-step explanation:

According to factor theorem, if x - 1 is a

factor of p(x), then p(1) = 0

(1) p(x) = x²+x+k

Since, x - 1 is a factor of the given polynomial p(x), thus p(1) = 0

→ P(1) - (1)² + (1) + k

→ 0=2+k

→ k = -2

(ii) p(x) = 2x² + kx + √2

Since, x - 1 is a factor of the given polynomial p(x), thus p(1) = 0 ·p(1) = 2(1)² + k(1) + √2

→ 0=2+k+ √2

→ k=-(2 + √2)

(III) p(x) = kx² -√2x+1

Since, x1 is a factor of the given

polynomial p(x), thus p(1) = 0

p(1) k(1)²-(√2 x 1) +1

0=k-√2+1

→ k = √2-1

(iv) p(x) = kx²-3x + k

Since, x - 1 is a factor of the given

polynomial p(x), thus p(1) = 0

→ p(1) k(1)2-3(1) + k

→ 0=2k-3

→ k = 3/2

Answer 2

Answer:

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how are you

can be friends ❤️

Step-by-step explanation:

actually I and my elder sister both of us use the same Id

so that time my sister has taken the Id