Question 1:
Find the value of the polynomial at \(5x-4x^2+3\) at(i)x = 0
(ii) x = −1
(iii)x = 2
Solution 1:\(\)
(i) \(p(x)=5x-4x^2+3\)
\(p(0) = 5(0)-4(0)^2+3\)
\( = 3\)
(ii)\(p(x) = 5x - 4x^2 + 3\)
\(p(-1)=5(-1)-4(-1)^2+3\)
\(= -5- 4(1)+3\)
\( = -6\)
(iii) \(p(x)= 5x -4x^2 +3\)
\(p(2) = 5(2) -4(2)^2+3 \)
\(= 10-16+3\)
\(=-3\)
Question 2:Find p(0), p(1) and p(2) for each of the following polynomials:
(i)\(p(y) = y^2 − y + 1\)
\(p(0) = (0)^2 − (0) + 1 \)
\(= 1\)
\(p(1) = (1)^2 − (1) + 1 \)
\(= 1 – 1 + 1\)
\( = 1\)
\(p(2) = (2)^2 − (2) + 1 \)
\(= 4 -2 +1 \)
\(= 3\)
(ii)\(p(t) = 2 + t + 2t^2 − t^3\)
\(p(0) = 2 + 0 + 2 (0)^2 − (0)^3 \)
\(= 2\)
\(p(1) = 2 + (1) + 2(1)^2 − (1)^3 \)
\(= 2 + 1 + 2 − 1 \)
\(= 4\)
\(p(2) = 2 + 2 + 2(2)^2 − (2)^3 \)
\(= 2 + 2 + 8 − 8 \)
\(= 4\)
(iii)\(p(x) = x^2\)
\(p(0) = (0)^3 = 0\)
\(p(1) = (1)^3 = 1\)
\(p(2) = (2)^3 = 8\)
\(p(0) = (0 − 1) (0 + 1) \)
\(= (− 1) (1) \)
\(= − 1\)
\(p(1) = (1 − 1) (1 + 1) \)
\(= 0 (2) \)
\(= 0\)
\(p(2) = (2 − 1 ) (2 + 1) \)
\(= 1(3) \)
\(= 3\)
Verify whether the following are zeroes of the polynomial, indicated against them.
(i)\(p(x) =3x +1, x =\dfrac{-1}{3}\)
Solution
If \(x = \dfrac{-1}{3}\) is a zero of given polynomial p\(\Bigg(\dfrac{-1}{3}\Bigg) = 3x + 1\), then, \(p\Bigg(\dfrac{-1}{3}\Bigg)\) should be \(0\).
Here, \(p(\dfrac{-1}{3}) = 3(\dfrac{-1}{3}) + 1 = -1+1 = 0\)
Therefore, \(p(x) =3x +1, x=\dfrac{-1}{3}\), is a zero of the given polynomial.
(ii)\(p(x) =5x -Ï€ ,x=\dfrac{4}{5}\)
Solution
If \(x=\dfrac{4}{5}\) is a zero of given polynomial
\(p(x) =5x-\pi\) then \(x=(\dfrac{4}{5})\) should be \(0\).
Here, \(p(\dfrac{4}{5})=5(\dfrac{4}{5})-\pi =4-\pi\).
As \(p(\dfrac{4}{5})≠0\).
Therefore, \(p(x) =5x-\pi \), is not a zero of the given polynomial.
(iii)\(p(x) =x^2-1, x=1,-1\)
Solution
If \(x = 1\) and \(x = −1\) are zeroes of polynomial \(p(x) = x^2 − 1\), then \(p(1)\) and \(p(−1)\) should be \(0\).
Here, \(p(1) = (1)^2 − 1 = 0\), and
\(p(−1) = (−1)^2 − 1 = 0\)
Hence, \(x = 1\) and \(−1\) are zeroes of the given polynomial.
(iv)\(p(x) =(x+1)(x-2), x=-1,2\)
Solution
If \(x = −1\) and \(x = 2\) are zeroes of polynomial \(p(x) = (x +1) (x − 2)\), then \(p(−1)\) and \(p(2)\) should be \(0\).
Here, \(p(−1) = (−1 + 1) (−1 − 2) = 0(−3) = 0\), and
\(p(2) = (2 + 1) (2 − 2) = 3(0) = 0\)
Therefore, \(x = −1\) and \(x = 2\) are zeroes of the given polynomial.
(v)\(p(x) =x^2, x=0\)
Solution
If \(x = 0\) is a zero of polynomial \(p(x) = x^2\), then \(p(0)\) should be zero.
Here, \(p(0) = (0)^2 = 0\)
Hence, \(x = 0\) is a zero of the given polynomial.
(vi)\(p(x) =lm+m, x=\dfrac{-m}{l}\)
Solution
p(x) = lx+m, x = −m/l
Solution:
For, \(x = {-m\over l} ; p(x) = lx+m\)
\(∴ p({-m\over l})= l({-m\over l})+m = −m+m = 0\)
\(∴ {-m\over l}\) is a zero of p(x).
Solution:
For, \(x = {-m\over l} ; p(x) = lx+m\)
\(∴ p({-m\over l})= l({-m\over l})+m = −m+m = 0\)
\(∴ {-m\over l}\) is a zero of p(x).
(vii)\(p(x) =3x^2-1, x=\dfrac{-1}{√3},\dfrac{-2}{√3}\)
Solution
\(p(x) = 3x^2−1, x = {-1\over \sqrt3} , {2\over\sqrt3}\)
Solution:
For, \(x = {-1\over \sqrt3} , {2\over\sqrt3} ; p(x) = 3x^2−1\)
Solution:
For, \(x = {-1\over \sqrt3} , {2\over\sqrt3} ; p(x) = 3x^2−1\)
\[∴ p({-1\over \sqrt3}) = 3({-1\over \sqrt3})2-1 = 3({1\over 3})-1 = 1-1 = 0\]
\[∴ p({2\over\sqrt3} ) = 3({2\over\sqrt3})2-1 = 3({4\over 3})-1 = 4−1 = 3 ≠ 0\]
\(∴ {-1\over \sqrt3}\) is a zero of p(x), but \({2\over\sqrt3}\) is not a zero of p(x).
(viii)\(p(x) =2x+1,x=\dfrac{1}{2}\)
Solution
\(p(x) =2x+1, x = {1\over 2}\)
Solution:
For, \(x = {1\over 2} p(x) = 2x+1\)
\(∴ p({1\over 2}) = 2({1\over 2})+1 = 1+1 = 2≠0\)
\(∴ {1\over 2}\) is not a zero of p(x).
Solution:
For, \(x = {1\over 2} p(x) = 2x+1\)
\(∴ p({1\over 2}) = 2({1\over 2})+1 = 1+1 = 2≠0\)
\(∴ {1\over 2}\) is not a zero of p(x).
Find the zero of the polynomial in each of the following cases:(i)p(x) = x + 5
(i)\(p(x) = x + 5\)
Let \(p(x) = 0\)
\(x + 5 = 0\)
\(x = − 5\)
Therefore, for \(x = −5\), the value of the polynomial is \(0\) and hence, \(x = −5\) is a zero of the given polynomial
(ii)p(x) = x – 5
Solution
Let \(p(x) = 0\)
\(x − 5 = 0\)
\(x = 5\)
Therefore, for \(x = 5\), the value of the polynomial is \(0\) and hence, \(x = 5\) is a zero of the given polynomial.
(iii)p(x) = 2x + 5
Solution
Let \(p(x) = 0\)
\(2x + 5 = 0\)
\(2x = − 5\)
\(x=\dfrac{-5}{2}\)
Therefore, \(x=\dfrac{-5}{2}\) for, the value of the polynomial is \(0\) and hence,\(x=\dfrac{-5}{2}\) is a zero of the given polynomial.
(iv) p(x) = 3x – 2
Solution
\(p(x) = 0\)
\(3x − 2 = 0\)
Therefore, for \(x=\dfrac{2}{3}\), the value of the polynomial is \(0\) and hence, \(x=\dfrac{2}{3}\) is a zero of the given polynomial.
(v)p(x) = 3x
\(3x − 2 = 0\)
Therefore, for \(x=\dfrac{2}{3}\), the value of the polynomial is \(0\) and hence, \(x=\dfrac{2}{3}\) is a zero of the given polynomial.
(v)p(x) = 3x
Solution
Let \(p(x) = 0\)
\(3x = 0\)
\(x = 0\)
Therefore, for \(x = 0\), the value of the polynomial is \(0\) and hence, \(x = 0\) is a zero of the given polynomial.
(vi) p(x) = ax, a ≠ 0
Solution
\(3x = 0\)
\(x = 0\)
Therefore, for \(x = 0\), the value of the polynomial is \(0\) and hence, \(x = 0\) is a zero of the given polynomial.
(vi) p(x) = ax, a ≠ 0
Solution
Let \(p(x) = 0\)
\(ax = 0\)
\(x = 0\)
Therefore, for \(x = 0\), the value of the polynomial is \(0\) and hence, \(x = 0\) is a zero of the given polynomial.
(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
Solution
Let \(p(x) = 0\)
\(cx + d = 0\)
\(x=\dfrac{-d}{c}\)
\(\therefore x=\dfrac{-d}{c}\) , the value of the polynomial is \(0\) and hence, \(x=\dfrac{-d}{c}\) is a zero of the given polynomial.
\(ax = 0\)
\(x = 0\)
Therefore, for \(x = 0\), the value of the polynomial is \(0\) and hence, \(x = 0\) is a zero of the given polynomial.
(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
Solution
Let \(p(x) = 0\)
\(cx + d = 0\)
\(x=\dfrac{-d}{c}\)
\(\therefore x=\dfrac{-d}{c}\) , the value of the polynomial is \(0\) and hence, \(x=\dfrac{-d}{c}\) is a zero of the given polynomial.
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