Question

find the zeroes of the following polynomials and verify the relationship between the zeroes and the coefficient x^2-12x+32
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Answer 1

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$\huge\tt{GIVEN:}$

• A polynomial x²-12x+32

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$\huge\tt{TO~FIND:}$

• The relationship between the zeros & the coefficients.

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$\huge\tt{SOLUTION:}$

f(x) = x² - 12x + 32

Coefficients,By Middle Term Factorisation,

↪ f(x) = x² - 8x - 4x + 32

↪ f(x) = x(x - 8) - 4(x - 8)

↪ f(x) = (x - 4)(x - 8)

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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

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$\huge\tt{VERIFICATION:}$

Let α and β be the zeroes of the polynomial. So, α = 4 and β = 8

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

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

→Sum of Zeroes = coefficient of x / coefficient of x²

→ Sum of Zeroes= α + β

→ Sum of Zeroes = α + β = 4 + 8 = 12

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

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→Product of Zeroes = Constant term / coefficient of x²

→Product of Zeroes = αβ = (4)(8)

→Product of Zeroes = αβ = 32

Also, c/a = 32/1 = 32

Hence, ✔️erified !!

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Answer 2

Answer :

The zeroes of the polynomial are 4 and 8

Given :

The quadratic polynomial is :

• x² - 12x + 32

Task :

• The zeroes of the given polynomial
• Also to verify the relationship between the zeroes and coefficients of the polynomial

Solution :

Factorizing the polynomial :

$\sf = x^{2} - 12x + 32 \\\\ \sf = x^{2} - 4x - 8x + 32 \\\\ \sf = x(x - 4) - 8 (x - 4) \\\\ \sf = (x - 4)(x - 8)$

Thus the zeroes are :

$\sf x - 4 = 0 \: \: and \: \: x - 8 = 0 \\\\ \sf \implies x = 4 \: \: and \: \implies x = 8$

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Verification of the relationship between zeroes and the coefficients :

$\sf \star \: \: Sum \: of \: the\: zeroes = -\dfrac{Coefficient \: of \: x}{Coefficient \: of \: x^{2}} \\\\ \sf \implies 4 + 8 = -\dfrac{-12}{1} \\\\ \sf \implies 12 = 12$

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$\sf \star \: \: Products \: \: of\: the \: zeroes = \dfrac{Constant \: \: term}{Coefficient \: of \: x^{2}} \\\\ \sf \implies 4\times 8 = \dfrac{32}{1} \\\\ \sf \implies 32 = 32$

$\bf Hence \: \: Verified$