Question

the zeros of polynomial x^2- 3x - 4 are

Answer 1

Answer:

x=1,-4

Step-by-step explanation:

x^2-3x-4

=> x^2-x+4x-4=0

=> x(x-1)+4(x-1)=0

=> (x-1)(x+4)=0

=>x= 1 or x= -4

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Answer 2

We have,

$\quad {x}^{2} - 3x - 4 \\ \\ = {x}^{2} - (4 - 1)x - 4 \\ \\ = {x}^{2} - 4x + x - 4 \\ \\ = x(x - 4) + 1(x - 4) \\ \\ = (x - 4)(x + 1)$

For finding the zeroes of the polynomial,

$\mathtt{let} \qquad {x}^{2} - 3x - 4 = 0 \\ \\ \implies (x - 4)(x + 1) = 0 \\ \\ \implies x - 4 = 0 \quad || \: or \quad x + 1 = 0 \\ \quad \quad \quad \quad \quad \: \: \quad \quad || \\ \implies x = 4 \qquad \: \: \: || \Rightarrow \: \: \: x = - 1$

Hence ( x = 4 ) or ( x = - 1 ) .

Thus, 4 and - 1 are the roots of the given polynomial.