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Citrus2011 [14]
2 years ago
13

A) x+4

Mathematics
1 answer:
KiRa [710]2 years ago
4 0

Answer:

  a) f(x) = x(x +4)(x -5)

  b) g(x) = (x -6)(x -2)(x^2 +2x +4)

  c) h(x) = (x +7)(x -3)(x -4)

Step-by-step explanation:

To factor completely generally means to find binomial factors that have integer constants. It is helpful to be familiar with the idea of "greatest common divisor", and what the product of two binomial factors looks like. In particular, there are a couple of special forms that can be useful. One of them is the difference of squares. Another is the difference of cubes.

  a² -b² = (a -b)(a +b)

  a³ -b³ = (a -b)(a² +ab +b²)

In the general case, the product of binomial factors is ...

  (x -a)(x -b) = x² -(a+b)x +ab

The coefficient of the x-terms is the sum of the binomial constants; the constant term is their product.

__

<h3>a)</h3>

The given cubic has no constant term. That is, every term has a factor of x, so that is one of the factors.

  f(x) = (x)(x^2 -x -20)

The given factor of x+4 tells you the constant in the other factor is -20/4 = -5.

  f(x) = x(x +4)(x -5)

__

<h3>b)</h3>

We observe that the ratios of pairs of coefficients are ...

  1 : -6 and -8 : 48 = 1 : -6 . . . . . pairs of coefficients have the same ratio

That means we can factor each pair of terms:

  g(x) = x^4 -6x^3 -8x +48

  g(x) = x^3(x -6) -8(x -6) = (x^3 -8)(x -6)

Using the difference of cubes formula, the factorization is ...

  g(x) = (x -2)(x^2 +2x +4)(x -6)

Note: the quadratic factor has irrational complex roots.

__

<h3>c)</h3>

The quadratic factor of this cubic can be found by dividing it by the given factor. That is most easily accomplished using synthetic division. Explaining how to do that is beyond the scope of this answer. The first attachment shows that the resulting factorization is ...

  h(x) = (x +7)(x^2 -7x +12)

The quadratic is factored by finding two factors of 12 that have a sum of -7. The positive sign of their product (12) tells you both will be negative.

  12 = (-1)(-12) = (-2)(-6) = (-3)(-4)

The pair of factors with a sum of -7 is -3 and -4. These are the constants in the remaining binomial factors:

  h(x) = (x +7)(x -3)(x -4)

_____

<em>Additional comment</em>

A graphing calculator can be an invaluable tool for factoring higher-degree polynomials. The graph of h(x) in the second attachment, for example, shows you the zeros are -7, 3, 4, so the factors are (x+7)(x-3)(x-4).

For a function like g(x), you can find the real zeros from the graph, then you can find the quadratic by factoring out the associated factors. The vertex of the quadratic curve helps you write the expression for that. (see the third attachment)

  g(x) = (x -2)(x -6)((x +1)^2 +3) . . . . quadratic factor in vertex form

  g(x) = (x -2)(x -6)(x^2 +2x +4)

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Assume that T is a linear transformation. Find the standard matrix of T. T: R^3 right arrow R^2 , T(e 1) =(1,2), and T(e2 ) =( -
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Answer:

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Step-by-step explanation:

Given

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Required

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T(x) = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]

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Ax = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]

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T(e_1) = (1,2)

T(e_2) = (-4,6)

T(e_3) = (2,-6)

In matrix:

(a,b) is represented as: \left[\begin{array}{c}a\\b\end{array}\right]

So:

T(e_1) = (1,2) = \left[\begin{array}{c}1\\2\end{array}\right]

T(e_2) = (-4,6)=\left[\begin{array}{c}-4\\6\end{array}\right]

T(e_3) = (2,-6)=\left[\begin{array}{c}2\\-6\end{array}\right]

Substitute the above expressions in A = [T(e_1)\ T(e_2)\ T(e_3)]

A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]

Hence, the standard of the matrix A is:

A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]

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