Question

Find the zeroes of the given polynomials.
(i) p(x) = 3x
(ii) p(x) = x square+ 5x + 6
(iii) p(x) = (x+2)(x+3)
(iv) p(x) = x power 4 - 16
its urgent please solve fast I want step by step please ​

Answer 1

◘ Question  ◘

Find the zeroes of the given polynomials.

(i) p(x) = 3x

(ii) p(x) = x² + 5x + 6

(iii) p(x) = (x + 2)(x + 3)

(iv) p(x) = x⁴ - 16

◘ We Must Know  ◘

To find the zero(s) of a given polynomial, it must be equated with zero. The possible values for 'x' will be the zeros of the polynomial.

◘ Solution  ◘

$\sf{(i)\ p(x)=3x}\\\\\sf{\dashrightarrow 3x=0}\\\\\sf{\dashrightarrow x=\dfrac{0}{3} }\\\\\sf{\color{magenta}{\dashrightarrow x=0}}$

Hence, the zero of this polynomial p(x) is 0.

$\rule{180}2$

$\sf{(ii)\ p(x)=x^2+5x+6}\\\\\sf{\dashrightarrow x^2+5x+6=0}\\\\\sf{\dashrightarrow x^2+2x+3x+6=0}\\\\\sf{\dashrightarrow x(x+2)+3(x+2)=0}\\\\\sf{\dashrightarrow (x+2)(x+3)=0}\\\\\sf{\dashrightarrow (x+2)=0\:\:or\:\:(x+3)=0}\\\\\sf{\color{magenta}{\dashrightarrow x=-2}\:\:or\:\:\color{magenta}{x=-3}}$

Hence, the zero of this polynomial p(x) are -2 & -3.

$\rule{180}2$

$\sf{(iii)\ p(x)=(x+2)(x+3)}\\\\\sf{\dashrightarrow (x+2)(x+3)=0}\\\\\sf{\dashrightarrow (x+2)=0\:\:or\:\:(x+3)=0}\\\\\sf{\color{magenta}{\dashrightarrow x=-2}\:\:or\:\:\color{magenta}{x=-3}}$

Hence, the zero of this polynomial p(x) are -2 & -3.

$\rule{180}2$

$\sf{(iv)\ p(x)=x^4-16}\\\\\sf{\dashrightarrow x^4-16=0}\\\\\sf{\dashrightarrow x^4=16}\\\\\sf{\dashrightarrow x=\sqrt[4]{16} }\\\\\sf{\color{magenta}{\dashrightarrow x=\pm 2}}$

Hence, the zeros of this polynomial p(x) are +2 & -2.

Answer 2

(i) p(x) = 3x

Now,

$3x=0$

$\boxed{\boxed{\bold{\implies x= 0}}}}}$

                                                                     

(ii) p(x) = x² + 5x + 6

Solving, we get,

$\implies x^{2} +2x+3x+6=0\\\implies x(x+2)+3(x+2)=0\\\implies (x+3)(x+2)=0$

Now,

$x+3=0$

$\boxed{\boxed{\bold{\implies x= -3}}}}}$

$x+2=0$

$\boxed{\boxed{\bold{\implies x= -2}}}}}$

                                                                     

(iii) p(x) = (x+2)(x+3)

$x+2=0$

$\boxed{\boxed{\bold{\implies x= -2}}}}}$

$x+3=0$

$\boxed{\boxed{\bold{\implies x= -3}}}}}$

                                                                       

(iv) p(x) = x⁴ - 16

$\implies (x^{2} -4)(x^{2} +4)\\\implies (x+2)(x-2)(x^{2} +4)$

$\boxed{\boxed{\bold{\implies x= -2, +2 }}}}}$