Find the zeroes of the given polynomials.(i) p(x) = 3x(ii) p(x) = x sq

1.	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 - 16its urgent please solve fast I want step by step please
Answer» ◘ 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.

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 - 16its urgent please solve fast I want step by step please

Answer»

◘ 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.