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2 years ago

(x+3)(x2 – 3x + 4) help simplify the expression.

Mathematics

2 answers:

MissTica2 years ago

5 0

Answer:

(x+3)(x2-3x+4)

x3-3x2+4x+3x2-9x+12=0

x3-3x2+3x2+4x-9x+12=0

x2-5x+12=0

x2-5x=-12

x2÷x-5x÷x=-12

1-5=-12

-4=-12

-4÷-4=-12÷-4

x=3

NNADVOKAT [17]2 years ago

4 0

Answer:

x³-5x+12

Step-by-step explanation:

(x+3)(x²-3x+4)

=x³-3x²+4x+3x²-9x+12

collect like terms

=x³-3x²+3x²+4x-9x+12

=x³-5x+12

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1. If a polynomial of Degree 4 has an imaginary zero at -2i and a real zero at 5, how many imaginary zeros does it have?

Vadim26 [7]

Answer:

Complex roots for polynomials equations will always come in conjugate pairs. Since 1-2i is a root, 1+2i will also be a root. (Just switch the sign of the imaginary part to its opposite.)

Step-by-step explanation:

So far we have 1-2i and 1+2i as roots. We need two more to make a 4th degree polynomial.

x=1 with multiplicity of 2 means that we count the root twice when we write down the polynomial: (x - 1)(x - 1) = (x - 1)2

Putting it all together, we can construct our polynomial of degree 4 as

y = f(x) = (x - 1)(x - 1)[x - (1-2i)][x - (1+2i)]

Multiply the factors (FOIL):

[x - (1-2i)][x - (1+2i)] =

x2 - (1+2i)x - (1-2i)x + (1-2i)(1+2i) =

x2 - x - 2ix - x + 2ix + (1 + 2i - 2i + 4) =

x2 - 2x + 5

(x - 1)(x - 1) = x2 - x - x + 1 = x2 - 2x + 1

Now multiply the two underlined expressions:

y = (x2 - 2x + 5)(x2 - 2x + 1)

= x4 - 2x3 + 5x2 - 2x3 + 4x2 - 10x + x2 - 2x + 5 by multiplying term by term

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3 years ago

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slavikrds [6]

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3 years ago

Please help solve this

Ne4ueva [31]

Answer:

last option

Step-by-step explanation:

We know that a₁ (or the first term) = 4 so we can rule out the second and third options. d (or the common difference) = 8 (because 4 + 8 = 12, 12 + 8 = 20 and so on) so our answer is the last option.

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3 years ago

Read 2 more answers

22. Determine the value of c so that the line

tiny-mole [99]

Answer:

If the lines BC and DE are parallel, the value of c is c=-2

Step-by-step explanation:

We are given line segment BC with end points B(2, 2) and C(9,6) and line segment DE with endpoints D(c, -7) and E(5, -3).

Using slope formula: $Slope=\frac{y_2-y_1}{x_2-x_1}$ we can find point c

When 2 lines are parallel there slope is same.

So, Slope of line BC =Slope of Line DE

$\frac{y_2-y_1}{x_2-x_1}=\frac{y_2-y_1}{x_2-x_1}$

We have:

$x_1=2, y_1=2, x_2=9, y_2=6 \ for \ line \ BC \ and \\\x_1=c, y_1=-7, x_2=5, y_2=-3 \ for \ line \ DE \$

Putting values and finding c

$\frac{6-2}{9-2}=\frac{-3-(-7)}{5-c}\\ \frac{4}{7}=\frac{-3+7}{5-c} \\ \frac{4}{7}=\frac{4}{5-c} \\Cross \ multiply:\\4(5-c)=4*7\\20-4c=28\\-4c=28-20\\-4c=8\\c=\frac{8}{-4}\\c=-2$

So, If the lines BC and DE are parallel, the value of c is c=-2

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3 years ago

(x+3)(x2 – 3x + 4) help simplify the expression.​

(x+3)(x2 – 3x + 4) help simplify the expression.