Question

Compute the zeroes of the polynomial 4x2 – 4x – 8. Also, establish a relationship between the zeroes and coefficients.​

Answer 1

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Let the given polynomial be $p(x) = 4x^2 – 4x – 8$

To find the zeroes, take $p(x) = 0$

Now, factorise the equation $4x^2 – 4x – 8 = 0$

$4x^2 – 4x – 8 = 0$

$4(x^2 – x – 2) = 0$

$x^2 – x – 2 = 0$

$x^2 – 2x + x – 2 = 0$

$x(x – 2) + 1(x – 2) = 0$

$(x – 2)(x + 1) = 0$

$x = 2, x = -1$

So, the roots of $4x^2 – 4x – 8 are -1 and 2.$

Relation between the sum of zeroes and coefficients:

$-1 + 2 = 1 = -(-4)/4(- coefficient \:of \:x/ coefficient \:of\: x^2)$

Relation between the product of zeroes and coefficients:

$(-1) × 2 = -2 =  -8/4 (constant/coefficient \:of \:x^2)$

Answer 2

Answer:

Step-by-step explanation:

4 {x}^{2}  - 4x - 8 = 0 \\ 4( {x}^{2}  - x - 2) = 0 \\  {x }^{2}  - 2x + x - 2 = 0 \\ x(x - 2) + 1(x - 2) = 0 \\ (x - 2)(x + 1) = 0 \\ x = 2 \: and \:  - 1 \\  

Sum of roots = - coefficient of x/coefficient of x^2

2+(-1)= - (-4)/4=1

Product of roots = constant /coefficient of x^2

2(-1)=-8/4= - 2

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