Polynomials — Question 3

(i) p(x) = 3x + 1

Putting x = $-\dfrac{1}{3}$ we get,

$\text{p}\Big(-\dfrac{1}{3}\Big)$ = $3\Big(-\dfrac{1}{3}\Big)$ + 1

= -1 + 1

= 0

Hence, x = $-\dfrac{1}{3}$ is zeroes of this polynomial.

(ii) p(x) = 5x - π

Putting x = $\dfrac{4}{5}$ we get,

$\text{p}\Big(\dfrac{4}{5}\Big) = 5\Big(\dfrac{4}{5}\Big) - \pi$

= 4 - π

Hence, x = $\dfrac{4}{5}$ is not a zeroes of this polynomial.

(iii) p(x) = x² - 1

Putting x = 1 we get,

p(1) = (1)² - 1

= 0

Putting x = -1 we get

p(-1) = (-1)² - 1

= 1 - 1

= 0

Hence, x = 1, -1 is zeroes of this polynomial.

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

Putting x = -1 we get,

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

= 0 x (-3)

= 0

Putting x = 2 we get,

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

= 3 x 0

= 0

Hence, x = -1, 2 are the zeroes of this polynomial.

(v) p(x) = x²

Putting x = 0 we get,

p(0) = (0)²

= 0

Hence, x = 0 is the zero of this polynomial.

(vi) p(x) = lx + m

Putting x = -$\dfrac{m}{l}$ we get,

p - $\Big(\dfrac{m}{l}\Big) = l x -\Big(\dfrac{m}{l}\Big)$ + m

= -m + m

= 0

Hence, x = $-\dfrac{m}{l}$ is the zero of this polynomial.

(vii) p(x) = 3x² - 1

Putting x = $-\dfrac{1}{\sqrt{3}}$ we get,

p-$\Big(\dfrac{1}{\sqrt{3}}\Big)$ = 3 x -$\Big(\dfrac{1}{\sqrt{3}}\Big)$² - 1

= 3 x $\dfrac{1}{3}$ - 1

= 1 - 1

= 0

Putting x = $\dfrac{2}{\sqrt{3}}$ we get,

p $\Big(\dfrac{2}{\sqrt{3}}\Big)$ = 3 x $\Big(\dfrac{2}{\sqrt{3}}\Big)$² - 1

= 3 x $\dfrac{4}{3}$ - 1

= 4 - 1

= 3

Hence, x = $-\dfrac{1}{\sqrt{3}}$ is zeroes of polynomial but x = $\dfrac{2}{\sqrt{3}}$ is not a zeroes of this polynomial.

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

Putting x = $\dfrac{1}{2}$ we get,

p$\Big(\dfrac{1}{2}\Big)$ = 2 x $\dfrac{1}{2}$ + 1

= 1 + 1

= 2

No, x = $\dfrac{1}{2}$ is not the zeroes of this polynomial.