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Find the value of the polynomial $P(x)= 5x-4x^2 +3$ at

(i) x = 0 (ii) x = -1 (iii) x = 2

(i)$P(x)= 5x-4x^2 +3$

P(0)= 0-0+3=3

(ii)$P(x)= 5x-4x^2 +3$

P(-1) =-5-4+3=-6

(iii)$P(x)= 5x-4x^2 +3$

P(2)= 10-16+3=-3

Find p(0), p(1) and p(2) for each of the following polynomials:

(i) $p(y) = y^2 - y + 1$

(ii) $p(t) = 2 + t + 2t^2- t^3$

(iii) $p(x) = x^3$

(iv) $p(x) = (x - 1) (x + 1)$

(i) $p(y) = y^2 - y + 1$

p(0) = (0)

p(1) = (1)

p(2) = (2)

(ii) $p(t) = 2 + t + 2t^2- t^3$>

p(0) = 2 + 0 + 2 (0)

p(1) = 2 + (1) + 2(1)

= 2 + 1 + 2 - 1 = 4

p(2) = 2 + 2 + 2(2)

= 2 + 2 + 8 - 8 = 4

(iii) $p(x) = x^3$

p(0) = (0)

p(1) = (1)

p(2) = (2)

(iv) $p(x) = (x - 1) (x + 1)$

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 =- 1/3

(ii) $p(x) = 5x - \pi $, x =4/5

(iii) $p(x) = x^2 - 1$, x = 1, -1

(iv) $p(x) = (x + 1) (x - 2)$, x = - 1, 2

(v) $p(x) = x^2$ , x = 0

(vi) $p(x) = lx + m$, x = - m/l

(vii) $p(x) = 3x^2 - 1$, x =-1/√3 and 2/√3

(viii) $p(x) = 2x + 1$, x =1/2

(i) $p(x) = 3x + 1$,x=-1/3

p(-1/3) = 3 (-1/3)+1=-1+1=0

p(-1/3) = 0 which means that-1 /3is zero of the polynomial p(x) = 3x+1.

(ii) $p(x) = 5x - \pi $, x =4/5

p(4/5) = 5(4/5)-π = 4-π

p(4/5) ≠ 0 which means that 4/5 is not zero of the polynomial $p(x) = 5x - \pi $.

(iii) $p(x) = x^2 - 1$,x = 1,- 1

p(1)=1

p(-1)=(-1)

Both p(1) and p(-1) are equal to 0. It means that 1 and -1 are zeroes of the polynomial $p(x) = x^2 - 1$.

(iv) $p(x) = (x + 1) (x - 2)$,x=-1,2

p(-1) = (-1+1)(-1-2) = 0×-3 = 0

p(2) = (2+1)(2-2) = 3×0 = 0

Both p(-1) and p(2) are equal to 0. It means that -1 and 2 are zeroes of the polynomial $p(x) = (x + 1) (x - 2)$.

(v) $p(x) = x^2$,x = 0

p(0) = 0

p(0) = 0 which means that 0 is the zero of the polynomial $p(x) = x^2$.

(vi) $p(x) = lx + m$,x =-m/

p(-m/l) =l(-m/l) + m = -m + m = 0

p(-m/l) = 0 which means that(-m/l)is zero of the polynomial $p(x) = lx + m$

(vii) $p(x) = 3x^2 - 1$, x =-1/√3 and 2/√3

p(-1/√3) = 3(-1/√3)

p(2/√3) = 3(2/√3)

p(-1/√3) = 0 which means that-1/√3is zero of the polynomial p(x) = 3x

p(2/√3) ≠ 0 which means that2/√3is not zero of the polynomial $p(x) = 3x^2 - 1$.

(viii) p(x) = 2x + 1 ,x =1/2

p(1/2)=2 ×(1/2)+ 1=1+1=2

p(1/2) ≠ 0. It means that 1/2 is not zero of the polynomial p(x)=2x+1.

Find the zero of the polynomial in each of the following cases:

(i) $p(x) = x + 5$

(ii) $p(x) = x - 5$

(iii) $p(x) = 2x + 5$

(iv) $p(x) = 3x - 2$

(v) $p(x) = 3x$

(vi) $p(x) = ax$, a ≠ 0

(vii) $p(x) = cx + d$, c ≠ 0, c, d are real numbers.

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

(i) $p(x) = x + 5$

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) $p(x) = x - 5$

p(x) = 0

x - 5 = 0

x = 5

Therefore, for x = 5, the value of the polynomial is0 and hence, x = 5 is a zero of the given polynomial.

(iii) $p(x) = 2x + 5$

p(x) = 0

2x + 5 = 0

2x = - 5

x = -5/2

Therefore, for x = -5/2 , the value of the polynomial is 0 and hence, x = -5/2is a zero of the given polynomial.

(iv) $p(x) = 3x - 2$

p(x) = 0

3x - 2 = 0

x=2/3, so 2/3 is the zero of the polynomial

(v) $p(x) = 3x$

p(x) = 0

3x = 0

x = 0

So x=0 is the zero of the polynomial

(vi) $p(x) = ax$

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) $p(x) = cx + d$

p(x) = 0

cx+ d = 0

x=-d/c

Download this assignment as pdf

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Given below are the links of some of the reference books for class 9 Math.

- Mathematics for Class 9 by R D Sharma One of the best book for studying class 9 level mathematics. It has lot of problems to be solved.
- Secondary School Mathematics for Class 9 by R S Aggarwal This is also as good as R.D. Sharma. Either this or the book by R.D. Sharma will do. I find book R.S. Aggarwal little bit more challenging than the one by R.D. Sharma.
- Pearson IIT Foundation Series - Maths - Class 9 Buy this book if you want to challenge yourself further and want to prepare for JEE foundation.
- Pearson IIT Foundation Physics, Chemistry & Maths combo for Class 9 Only buy if you are prepared to study extra topics and want to take your studies a step further. You might need help to understand topics in these books.

You can use above books for extra knowledge and practicing different questions.

Class 9 Maths Class 9 Science

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