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3 years ago

Express the polynomial as a product of linear factors. 3x^3+12x^2+3x-18

Mathematics

2 answers:

Marysya12 [62]3 years ago

8 0

Answer:

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

Step-by-step explanation:

A logical first step would be to factor 3 out of all four terms:

f(x) = 3x^3+12x^2+3x-18 = 3(x^3 + 4x^2 + x - 6)

Roots of this x^3 + 4x^2 + x - 6 could be factors of 6: {±1, ±2, ±3, ±6}.

I would use synthetic division here to determine which, if any, of these possibilities are actually roots of x^3 + 4x^2 + x - 6. Let's try x = 1 and see whether the remainder of this synth. div. is 0, which would indicate that 1 is indeed a root of x^3 + 4x^2 + x - 6:

1 / 1 4 1 -6

1 5 6

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

1 5 6 0

Yes, 1 is a root of x^3 + 4x^2 + x - 6, and so (x - 1) is a factor of x^3 + 4x^2 + x - 6.

Look at the coefficients of the quotient, which are 1, 5 and 6.

This represents the quadratic 1x² + 5x + 6, whose factors are (x + 2) and

(x + 3).

Thus, the given polynomial in factored form is:

f(x) = 3x^3+12x^2+3x-18 = 3(x + 2)(x - 1)(x + 3)

NeX [460]3 years ago

3 0

Answer:

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

Step-by-step explanation:

apex

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Solve the system of equations 2x + 2y = 6 and 2y - x = 12 using any method.

beks73 [17]

2y + 2x = 6
2y - x = 12
Subtract
3x = -6, x = -2
2(-2) + 2y = 6
-4 + 2y = 6
2y = 10, y = 5
Final solution: x = -2, y = 5

8 0

2 years ago

Read 2 more answers

A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si

Grace [21]

Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

Let the dimension of the paper be x and y;

Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

Differentiate A'

$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

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larisa [96]

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Kitty [74]

Answer:

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