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3 years ago

We want to find the zeros of this polynomial p(x)= ( x^2-1)(x^2-5x+6)

Mathematics

1 answer:

Scrat [10]3 years ago

4 0

Answer:

see explanation

Step-by-step explanation:

To find the zeros let p(x) = 0 , that is

(x² - 1)(x² - 5x + 6) = 0

Factorise each factor

x² - 1 ← is a difference of squares and factors as (x - 1)(x + 1)

x² - 5x + 6 = (x - 2)(x - 3), thus

(x - 1)(x + 1)(x - 2)(x - 3) = 0

Equate each factor to zero and solve for x

x - 1 = 0 ⇒ x = 1

x + 1 = 0 ⇒ x = - 1

x - 2 = 0 ⇒ x = 2

x - 3 = 0 ⇒ x = 3

The zeros are x = ± 1, x = 2, x = 3

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Read 2 more answers

Solve. -7x=x^2+12 help

Nat2105 [25]

Hey There!

The answer to your problem is $x=-3$ or $x=-4$ both answers are correct!

Step 1: Subtract x²+12 from both sides.

−7x−(x²+12)=x²+12−(x²+12)

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