Question

Q2. a) Find the zeroes of the
quadratic polynomial 4y2 +8y and
verify the relationship between the
zeroes and the coefficients. (2 marks)​

Answer 1

4y2+8y

4y(y+2)

therefore:

>4y=0. >y+2=0

y=0/4. y=0-2

y=0. y= -2

so, zeroes are 0 and -2

Answer 2

Answer:

0, -2 are the zeroes of the quadratic polynomial ($4y^2 + 8y$).

Step-by-step explanation:

Explanation:

Given in the question that, a quadratic polynomial $4y^2 + 8y$.

• Quadratic Polynomial - A quadratic polynomial is one that has the form a$x^2$+bx + c, where a, b, and c are real numbers and a ≠ 0.

• Zeroes of a quadratic polynomial, each and every quadratic polynomial has two zeros.

Step 1:

We have $4y^2 + 8y$.

Taking 4y as common from the  given quadratic polynomial,

⇒4y (y + 2)

So, the zeroes of the polynomial are,

4y = 0 and y + 2 = 0

⇒ y = 0 and y = -2 .

Step 2:

4 is the coefficient of $y^2$  and 8 is the coefficient of y.

It means a = 4 , b = 8 and c = 0.

Relation between the zeroes and the coefficients;

sum of zeroes  are  $\frac{-b}{a}$ = $\frac{-8}{4}$ = -2

and product of zeroes are $\frac{c}{a}$ = 0

Final answer:

Hence, 0 and -2 are the zeroes of the given quadratic polynomial.

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