Question

1. Find the zeroes of the following quadratic
polynomial and verify the relationship between
the zeroes and its coefficients.
x2 – 2x–8​

Answer 1

Answer:

Hey mats heres your answer

Step-by-step explanation:

X^2 - (4-2)x - 8

X^2 - 4x - 2x - 8

X(x - 4) - 2(x - 4)

(X - 4) (x +2)

X - 4 = 0

X =4

X + 2 = 0

X = - 2

Therefore the zeros of the polynomial are - 2 and 4

Answer 2

ANSWER:

• Zeros of polynomial are 4 and (-2)

GIVEN:

• P(x) = x²-2x-8

TO VERIFY:

• Relationship between zeros and coefficients

SOLUTION:

=> x²-2x-8 =0

=> x²-4x+2x -8 = 0

=> (x²-4x) +(2x-8) = 0

=> x(x-4) +2(x-4) = 0

=>(x-4)(x+2) = 0

Either (x-4) = 0

=> x-4 = 0

=> x = 4

Either (x+2) = 0

=> x+2 = 0

=> x = (-2)

Now

Sum of zeros:

$= 4 + ( - 2) \\ \\ \implies \: \frac{ - ( - 2)}{1} = \dfrac{ - (coefficient \: of \: x)}{ coefficient \: of \: x {}^{2} } \:$

Product of zeros:

$= 4 \times ( - 2) \\ = - 8 \\ \\ \implies \: \frac{ - 8}{1} = \frac{constant \: term}{coefficient \: of \: x {}^{2} }$

• Zeros of polynomial are 4 and (-2)