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2 years ago

Select all the factors roots of x^3 +2x^2-16x-32

Mathematics

2 answers:

Ivanshal [37]2 years ago

5 0

Factor by grouping. Group up the terms into pairs, factor each pair, then factor out the overall GCF.

x^3 + 2x^2 - 16x - 32

(x^3 + 2x^2) + (-16x-32) ... pair up terms

x^2(x + 2) + (-16x - 32) ... factor x^2 from the first group

x^2(x + 2) - 16(x + 2) ... factor -16 from the second group

(x^2 - 16)(x + 2) .... factor out (x+2)

(x - 4)(x + 4)(x + 2) .... Use the difference of squares to factor x^2-16

---------------------------

The original expression completely factors to (x - 4)(x + 4)(x + 2)

The three factors are x - 4 and x + 4 and x + 2

solmaris [256]2 years ago

5 0

The answers are b) -2, c) -4, and e) 4.

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find the centre and radius of the following Cycles 9 x square + 9 y square +27 x + 12 y + 19 equals 0

Citrus2011 [14]

Answer:

Radius: $r =\frac{\sqrt {21}}{6}$

$Center = (-\frac{3}{2}, -\frac{2}{3})$

Step-by-step explanation:

Given

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Solving (a): The radius of the circle

First, we express the equation as:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

So, we have:

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Divide through by 9

$x^2 + y^2 + 3x + \frac{12}{9}y + \frac{19}{9} = 0$

Rewrite as:

$x^2 + 3x + y^2+ \frac{12}{9}y =- \frac{19}{9}$

Group the expression into 2

$[x^2 + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}$

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

Next, we complete the square on each group.

For $[x^2 + 3x]$

1: Divide the $coefficient\ of\ x\ by\ 2$

2: Take the $square\ of\ the\ division$

3: Add this $square\ to\ both\ sides\ of\ the\ equation.$

So, we have:

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

$[x^2 + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Factorize

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Apply the same to y

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}$

Add the fractions

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}$

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

By comparison:

$r^2 =\frac{7}{12}$

Take square roots of both sides

$r =\sqrt{\frac{7}{12}}$

Split

$r =\frac{\sqrt 7}{\sqrt 12}$

Rationalize

$r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}$

$r =\frac{\sqrt {84}}{12}$

$r =\frac{\sqrt {4*21}}{12}$

$r =\frac{2\sqrt {21}}{12}$

$r =\frac{\sqrt {21}}{6}$

Solving (b): The center

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

From:

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

$-h = \frac{3}{2}$ and $-k = \frac{2}{3}$

Solve for h and k

$h = -\frac{3}{2}$ and $k = -\frac{2}{3}$

Hence, the center is:

$Center = (-\frac{3}{2}, -\frac{2}{3})$

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2 years ago

2) Fill in an angle measure greater than 100 degrees for the missing angle in yellow below. You will fill

timama [110]

Answer:

m∠a = 110°

m∠b = 110°

m∠c = 70°

m∠d = 70°

m∠i = 110°

m∠h = 110°

m∠j = 70°

m∠g = 110°

m∠f = 70°

m∠e = 70°

m∠k = 57°

m∠m = 70°

m∠l = 167°

Step-by-step explanation:

Let the measure of the yellow angle = 110°

Therefore;

m∠a = 110° by vertically opposite angles

m∠b = m∠a = 110° by opposite interior angles of a parallelogram

m∠c = 180° - 110° = 70° by same side interior angles

m∠d = m∠c = 70° by opposite interior angles of a parallelogram

m∠i and m∠c are supplementary angles, therefore, m∠i = 110°

m∠h = m∠i = 110° by vertically opposite angles

m∠j = m∠c = 70° by vertically opposite angles

m∠g and m∠d are supplementary angles, therefore, m∠g = 110°

m∠f = m∠d = 70° by alternate interior angles

m∠f = m∠e = 70° by vertically opposite angles

m∠k = 180° - (m∠h + 13°) = 180° - (110° + 13°) = 57°

m∠b and m∠m are supplementary angles, therefore, m∠m = 70°

m∠l and 13° are supplementary angles, therefore, m∠l = 167°

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2 years ago

The measure of EFH is twice the measure of GFH. What is the

Vladimir [108]

Answer:

30° = m∠GFH

Step-by-step explanation:

Since ∠EFH and ∠GFH are complementary, the sum of their measures i 90.

Let x = m∠GFH

2x = m∠EFH

m∠GFH + m∠EFH = 90

x + 2x = 90

3x = 90

x = 30° = m∠GFH

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kozerog [31]

Answer:

Step-by-step explanation:

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