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05/07/2021 12:18 pm
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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 + √2
(iii) p(x) = kx2–√2x + 1
(iv) p(x) = kx2 – 3x + k
1 Answer
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05/07/2021 12:19 pm
(i) p(x) = x2+x+k
If x-1 is a factor of p(x), then p(1) = 0
By Factor Theorem
⇒ (1)2+(1)+k = 0
⇒ 1+1+k = 0
⇒ 2+k = 0
⇒ k = −2
(ii) p(x) = 2x2+kx+√2
If x-1 is a factor of p(x), then p(1)=0
⇒ 2(1)2+k(1)+√2 = 0
⇒ 2+k+√2 = 0
⇒ k = −(2+√2)
(iii) p(x) = kx2–√2x+1
If x-1 is a factor of p(x), then p(1)=0
By Factor Theorem
⇒ k(1)2-√2(1)+1=0
⇒ k = √2-1
(iv) p(x)=kx2–3x+k
If x-1 is a factor of p(x), then p(1) = 0
By Factor Theorem
⇒ k(1)2–3(1)+k = 0
⇒ k−3+k = 0
⇒ 2k−3 = 0
⇒ k= 3/2