Question

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) p(x)=2x²+kx+ $\sqrt{2}$
(iii) p(x)=kx²−$\sqrt{2}$x+1

Answer 1

(i) p(x)=x²+x+k

Zero of (x−1) is 1.

It is given that x−1 is a factor of p(x).

Therefore, by Factor theorem, p(1)=0.

p(1) =$1^2$+1+k=0

1+1+k=0

2+k=0

k= −2

(ii) p(x)=2x²+kx+$\sqrt{2}$

Zero of (x−1) is 1.

It is given that x−1 is a factor of p(x).

Therefore, by Factor theorem, p(1)=0.

p(1)=2(1)²+k(1)+ $\sqrt{2}$=0

2+k+$\sqrt{2}$=0

k=−2−$\sqrt{2}$

(iii) p(x)=kx²−$\sqrt{2}$x+1

Zero of (x−1) is 1.

It is given that x−1 is a factor of p(x).

Therefore, by Factor theorem, p(1)=0.

p(1)=k${\left(1\right)}^2$− $\sqrt{2}$ (1)+1=0

k− $\sqrt{2}$+1=0

k= $\sqrt{2}$ −1