Mathematics: ax+b=cx+b where a, b,c are constants

08.04.2021

ax+b=cx+b
where a, b,c are constants

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Mathematics

The polynomial (2x - 1)(x2 – 2) - x(x2 - x - 2) can be written in the form ax3 + bx2 + cx + d, where a, b, c, and d are constants. What are the values of a, b, c, and d?

Step-by-step answer

P Answered by PhD

On solving the polynomial

(2x-1)(x^2-2)-x(x^2-x-2)

=2x^3-4x-x^2+2-x^3+x^2+2x

=x^3-2x+2

on comparing this polynomial with ax^3 + bx^2 + cx + d

a=1, b=0, c=-2 and d=2

Mathematics

If f(x) = ax^6 + bx^4 + cx^3 where a,b, and c are integers, how many distinct rational zeros could f(x) have

Step-by-step answer

P Answered by Specialist

You can rewrite your function as

f(x) = x^3(ax^3+bx+c)

This implies that

$f(x)=0 \iff x^3=0\quad\lor\quad ax^3+bx+c=0$

Now, we have $x^3=0\iff x=0$ , so it counts as a solution.

On the other hand, depending on the coefficient a, b and c, the cubic equation

ax^2+bx+c=0

can have either one or three solutions.

So, we have the solution x=0, and then one or three solutions coming from the cubic part. The equation as a whole thus have either two or four solutions, depending on the coefficients.

Mathematics

show that (x+1)(x+3)(x+5) can be written in the form ax^3 + bx^2 + cx+ d where a,b,c and d are all positive integers.

Step-by-step answer

P Answered by PhD

x³ + 9x² + 23x + 15

Step-by-step explanation:

Given

(x + 1)(x + 3)(x + 5) ← expand the first pair of factors using FOIL

= (x² + 4x + 3)(x + 5)

Each of the terms in the second factor is multiplied by each of the terms in the first factor, that is

x²(x + 5) + 4x(x + 5) + 3(x + 5) ← distribute the 3 parenthesis

= x³ + 5x² + 4x² + 20x + 3x + 15 ← collect like terms

= x³ + 9x² + 23x + 15 ← in the form ax³ + bx² + cx + d

Mathematics

Show that (x+1)(x+3)(x+5) can be written in the form ax^3+bx^2+cx+d where a,b and c are all positive integers, and d is a negative integer

Step-by-step answer

P Answered by Specialist

x^3+9x2+23x+15

Step-by-step explanation:

Just Expand the brackets

(x^2+4x+3)(x+5)

x^3+9x2+23x+15

Hope you understood(give me brainliest answer) ;)

Mathematics

Show that (x-1)(x+2)(x+3) can be written in the form ax^3+bx^2+cx+d where a,b and c are all positive integers, and d is a negative integer

Step-by-step answer

P Answered by PhD

The answer is x³ + 4x² + x - 6.

Step-by-step explanation:

You have to expand it :

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

$= ( {x}^{2} + 2x - x - 2)(x + 3)$

$= ( {x}^{2} + x - 2)(x + 3)$

$= {x}^{3} + 3 {x}^{2} + {x}^{2} + 3x - 2x - 6$

$= {x}^{3} + 4 {x}^{2} + x - 6$

Mathematics

Show that (x+1)(x+3)(x+5) can be written in the form ax^3+bx^2+cx+d where a,b and c are all positive integers, and d is a negative integer

Step-by-step answer

P Answered by Specialist

x^3+9x2+23x+15

Step-by-step explanation:

Just Expand the brackets

(x^2+4x+3)(x+5)

x^3+9x2+23x+15

Hope you understood(give me brainliest answer) ;)

Mathematics

Show that (x+1)(x+3)(x+5) can be written in the form ax^3+bx^2+cx+d where a,b and d are all positive integers

Step-by-step answer

P Answered by PhD

(x+1) (x+3)(x+5)

(x^2+4x+3)(x+5)

x^3+4x^2+3x+5x^2+20x+15

x^3+9x^2+23x+15

1st you have to do foil method with the two first quantities and 2nd do foil with last quantity.
:)

Mathematics

Show that (x-1)(x+2)(x+3) can be written in the form ax^3+bx^2+cx+d where a,b and c are all positive integers, and d is a negative integer

Step-by-step answer

P Answered by PhD

The answer is x³ + 4x² + x - 6.

Step-by-step explanation:

You have to expand it :

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

$= ( {x}^{2} + 2x - x - 2)(x + 3)$

$= ( {x}^{2} + x - 2)(x + 3)$

$= {x}^{3} + 3 {x}^{2} + {x}^{2} + 3x - 2x - 6$

$= {x}^{3} + 4 {x}^{2} + x - 6$

Mathematics

If f(x) = ax^6 + bx^4 + cx^3 where a,b, and c are integers, how many distinct rational zeros could f(x) have

Step-by-step answer

P Answered by Specialist

You can rewrite your function as

f(x) = x^3(ax^3+bx+c)

This implies that

$f(x)=0 \iff x^3=0\quad\lor\quad ax^3+bx+c=0$

Now, we have $x^3=0\iff x=0$ , so it counts as a solution.

On the other hand, depending on the coefficient a, b and c, the cubic equation

ax^2+bx+c=0

can have either one or three solutions.

So, we have the solution x=0, and then one or three solutions coming from the cubic part. The equation as a whole thus have either two or four solutions, depending on the coefficients.

Mathematics

Show that (x+1)(x+3)(x+5) can be written in the form ax^3+bx^2+cx+d where a,b and d are all positive integers

Step-by-step answer

P Answered by PhD

1st you have to do foil method with the two first quantities and 2nd do foil with last quantity.
:)

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