find the value of p for which the polynomial 2 x ki power 4 + px ki power 3 +4x ki power 2 + 2x + 1 is exactly divoded by x-1/2
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i) x2 – 2x – 8
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u
(v) t2 – 15
(vi) 3x2 – x – 4
Sol. (i) x2 – 2x – 8
We have p(x) = x2 – 2x – 8
= x2 + 2x – 4x – 8 = x (x + 2) – 4 (x + 2)
= (x – 4) (x + 2)
For p(x) = 0, we have
(x – 4) (x + 2) = 0
Either x – 4 = 0 ⇒ x = 4
or x + 2 = 0 ⇒ x = – 2
∴ The zeroes of x2 2x – 8 are 4 and –2.
Now, sum of the zeroes 


Thus, relationship between zeroes and the coefficients in x2 – 2x – 8 is verified.
(ii) 4s2 – 4s + 1
We have p(s) = 4s2 – 4s + 1
= 4s2 – 2s – 2s + 1 = 2S (2S – 1) –1 (2s – 1)
= (2s – 1) (2s – 1)
For p(s) = 0, we have,

Now,
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u
(v) t2 – 15
(vi) 3x2 – x – 4
Sol. (i) x2 – 2x – 8
We have p(x) = x2 – 2x – 8
= x2 + 2x – 4x – 8 = x (x + 2) – 4 (x + 2)
= (x – 4) (x + 2)
For p(x) = 0, we have
(x – 4) (x + 2) = 0
Either x – 4 = 0 ⇒ x = 4
or x + 2 = 0 ⇒ x = – 2
∴ The zeroes of x2 2x – 8 are 4 and –2.
Now, sum of the zeroes 


Thus, relationship between zeroes and the coefficients in x2 – 2x – 8 is verified.
(ii) 4s2 – 4s + 1
We have p(s) = 4s2 – 4s + 1
= 4s2 – 2s – 2s + 1 = 2S (2S – 1) –1 (2s – 1)
= (2s – 1) (2s – 1)
For p(s) = 0, we have,

Now,
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