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Zeros of a Polynomial

1. Two zeroes...

Question

1.Two zeroes of the polynomial p(x)=x cube-4x square+x+6 are 2 and -1 of p(x)

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Solution

From given equation
a=1
b=-4
c=1
d=6
given two zeros -1 and 2
We know sum of zeros should be= -b/a
so z1+z2+z3= -(-4)/1
-1+2+z3=4
So z3=3

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Similar questions

Q. Verify that:
(i) 4 is a zero of the polynomial p(x) = x − 4.
(ii) −3 is a zero of the polynomial p(x) = x − 3.
(iii)

- \frac{1}{2}

is a zero of the polynomial p(y) = 2y + 1.
(iv)

\frac{2}{5}

is a zero of the polynomial p(x) = 2 − 5x.
(v) 1 and 2 are the zeros of the polynomial p(x) = (x − 1)(x − 2).
(vi) 0 and 3 are the zeros of the polynomial p(x) = x² − 3x.
(vii) 2 and −3 are the zeros of the polynomial p(x) = x² + x − 6.

Q. Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1, x =

- \frac{1}{3}

p (x) = 5 x - π, x = \frac{4}{5}

p (x) = x^{2} - 1, x = 1, - 1

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

Q. Question 3
Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1, x =

- \frac{1}{3}

p (x) = 5 x - π, x = \frac{4}{5}

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

p (x) = i x + m, x = - \frac{m}{i}

p (x) = 3 x^{2} - 1, x = - \frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}

p (x) = 2 x + 1, x = \frac{1}{2}

x - 2

x - 1

are the only zeroes of the polynomial

p (x)

, which of the following is a zero of the polynomial

p (x) - 6

?

Q. Question 3
Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1, x =

- \frac{1}{3}

p (x) = 5 x - π, x = \frac{4}{5}

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

p (x) = i x + m, x = - \frac{m}{i}

p (x) = 3 x^{2} - 1, x = - \frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}

p (x) = 2 x + 1, x = \frac{1}{2}

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