Question

Find the zeroes of the quadratic polynomial x^2-12x+32 and verify the relationship between the zeroes and the coefficients​

Answer 1

Answer :

Zeroes are 4 and 8.

Step-by-step explanation :

f(x) = x² - 12x + 32

By Middle Term Factorisation

→ f(x) = x² - 8x - 4x + 32

→ f(x) = x(x - 8) - 4(x - 8)

→ f(x) = (x - 4)(x - 8)

To find zeroes, f(x) = 0, then

→ 0 = (x - 4)(x - 8)

By Zero Product Rule

→ x - 4 = 0 and x - 8 = 0

→ x = 4 and x = 8

Let α and β be the zeroes of the polynomial.

So, α = 4 and β = 8

Verification :

On comparing the given polynomial with ax² + bx + c, we get

a = 1, b = - 12, c = 32

• Sum of Zeroes :

→ α + β = 4 + 8 = 12

Also, - b/a = - (- 12)/1 = 12

• Product of Zeroes :

→ αβ = (4)(8) = 32

Also, c/a = 32/1 = 32

Hence, verified !!

Answer 2

Answer:

8 and 4 are the zeroes of the polynomial.

Step-by-step explanation:

Explanation:

Given in the question that, $x^2 - 12x +32$

• The locations where a polynomial becomes zero overall are known as the zeros of the polynomial.

• The term "zero polynomial" refers to a polynomial with zero value (-1).

Step 1:

We have, $x^2 - 12x +32$

⇒ $x^2 - 8x -4x + 32$

⇒x(x - 8) + 4(x - 8)

⇒(x - 8) (x - 4)

Now, x - 8 = 0 ⇒ x = 8

and x - 4 = 0 ⇒ x = 4

So, 8 and 4 are the zeroes of the polynomial.

Step 2:

From step 1 we have, 8 and 4 are the zeroes of the polynomial.

And from the question we have, $x^2 - 12x +32$

Now as we know that,

Sum of zeroes ($\alpha + \beta$) = $\frac{-b}{a}$ and

Product of zeroes ($\alpha \beta$ ) = $\frac{c}{a}$

Where, $\alpha \ and \beta$ are the zeroes of the polynomial

And from the question we have, a = 1 , b = -12 and c = 32.

Therefore,

$(\alpha + \beta )$ = $\frac{-b}{a}$

⇒ (8 + (4)) = $\frac{12}{1}$

⇒ 12 = 12

Now, a product of zeroes ($\alpha \beta$) = $\frac{c}{a}$

⇒ 8 × 4 = $\frac{32}{1}$

⇒ 32 = 32

Here, we verified the relationship between the zeroes and the coefficient

Final answer:

Hence, 8 and 4 are the zeroes of the polynomial.

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