Question

obtain all zeros of the polynomial f of x is equal to x cube + 13x^2 +32x +20 if one of its zeros are -2
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Answer 1

Question:

Obtain all zeros of the polynomial f(x) = x³ + 13x² + 32x + 20 if one of its zeroes are -2.

Given:

f(x) = x³ + 13x² + 32x + 20

One of it's zero is -2, let us assume that x is the zero.

⇒ x = -2

To find:

All the other zeroes of x³ + 13x² + 32x + 20 other than -2.

Solution:

We've been given that -2 is one of the zeroes of the given polynomial.

$\Longrightarrow \sf x = -2\\ \\ \sf \Longrightarrow x + 2 = 0\\ \\ \therefore \underline{\underline{\textbf{(x + 2)} \sf \ is \ a \ factor \ of \ the \ polynomial \ f(x) = x^3 + 13x^2 + 32x + 20}}$

Now, we divide f(x) by (x + 2), and factorize the quotient to obtain the other zeroes.

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(Refer to the attachment for the division)

• Divisor = (x + 2)

• Dividend = x³ + 13x² + 32x + 20

• Quotient = x² + 11x + 10

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Now,

$\Longrightarrow \tt x^2 + 11x + 10\\ \\ \\\boxed{\begin{minipage}{4 cm}\sf \ \ \ \ \ \ \ \ \ Sum \dashrightarrow \sf 10 \\ \\\sf {\ \ \ \ \ \ \ Product \dashrightarrow 11}\\ \\\sf { \ \ \ \ \ \ \ Split \dashrightarrow 10 \times 1}\end{minipage}}\\ \\ \\ \\\Longrightarrow \tt x^2 + 10x + x + 10\\ \\ \Longrightarrow \tt x(x + 10) + 1(x + 10)\\ \\ \Longrightarrow \tt (x + 10) (x + 1)$

Equating the above with zero we get,

$\Longrightarrow \sf x + 10 = 0\\ \\ \Longrightarrow \sf x = -10\\ \\ \\\Longrightarrow \sf x + 1 = 0\\ \\ \Longrightarrow \sf x = -1$

Therefore, all the zeroes of the polynomial f(x) = x³ + 13x² + 32x + 20 are -2, -10 & -1.

Answer 2

Answer:

all the zeroes of the polynomial f(x) = x³ + 13x² + 32x + 20 are -2, -10 & -1. ...

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