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For the polynomial below, 1 is a zero.

h(x)=x ^3+ 5x^2+ x − 7

Express h (x) as a product of linear factors.

h(x)=

For The Polynomial Below 1 Is A Zero Hxx 3 5x2 X 7 Express H X As A Product Of Linear Factors Hx class=

Sagot :

sqdancefan

Answer:

  h(x) = (x -1)(x +3+√2)(x +3-√2)

Step-by-step explanation:

You want the linear factors of h(x) = x^3 +5x^2 +x -7, given that 1 is a zero.

Quotient

If 1 is a zero, then (x -1) is a factor. The remaining quadratic factor can be found using synthetic division, as shown in the attachment. This tells us the factorization is ...

  h(x) = (x -1)(x^2 +6x +7)

Quadratic factors

The factors of the quadratic can be found by completing the square:

  x^2 +6x = -7

  x^2 +6x +9 = 2 . . . . . . . add (6/2)^2 = 9 to both sides

  (x +3)^2 = 2 . . . . . . . . show as a squre

  x +3 = ±√2 . . . . . . . . take the square root

  x +3 ±√2 = 0 . . . . . . subtract ±√2; the two remaining factors

This tells us that h(x) = 0 when (x +3 +√2) = 0 and when (x +3 -√2) = 0. These binomial expressions are the remaining linear factors of h(x).

  h(x) = (x -1)(x +3+√2)(x +3-√2)

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