Junior Cycle Mathematics ALGEBRA - Expressions: Divide $x^3 + 5x^2 - 29x

Divide $x^3 + 5x^2 - 29x - 105$ by $x + 3$. - Junior Cycle Mathematics - Question b

Question b

Divide $x^3 + 5x^2 - 29x - 105$ by $x + 3$.

Worked Solution & Example Answer:Divide $x^3 + 5x^2 - 29x - 105$ by $x + 3$. - Junior Cycle Mathematics - Question b

Step 1

Long Division Setup

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Answer

To divide the polynomial $x^3 + 5x^2 - 29x - 105$ by $x + 3$ , we set up the long division as follows:

       x^2 + 2x - 35
     ______________________
x + 3 | x^3 + 5x^2 - 29x - 105

Here, we will divide $x^3$ by $x$ .

Step 2

First Division

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Answer

The first term of the quotient is $x^2$ , since:

$rac{x^3}{x} = x^2$

We then multiply $x^2$ by $(x + 3)$ and subtract:

$x^3 + 3x^2$

Step 3

Subtracting the First Product

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Answer

Subtracting gives:

&(x^3 + 5x^2) - (x^3 + 3x^2) \\ &= (5x^2 - 3x^2) = 2x^2 \end{aligned}$$ So we have: $$2x^2 - 29x$$

Step 4

Next Terms Division

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Answer

Next, bring down the $-105$ to get:

$2x^2 - 29x - 105$

Now divide $2x^2$ by $x$ to get $2x$ .

Step 5

Multiply and Subtract Again

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Answer

Now multiply $2x$ by $(x + 3)$ :

$2x^2 + 6x$

Subtracting again:

$(2x^2 - 29x) - (2x^2 + 6x) = -35x$

Now we have:

$-35x - 105$

Step 6

Final Division

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Answer

Next, divide $-35x$ by $x$ to get $-35$ .

Multiply $-35$ by $(x + 3)$ :

$-35x - 105$

Subtract to get:

$0$

Step 7

Final Result

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Answer

The division completes with a remainder of $0$ :

Therefore, the quotient is:

$x^2 + 2x - 35$

Divide $x^3 + 5x^2 - 29x - 105$ by $x + 3$. - Junior Cycle Mathematics - Question b

Worked Solution & Example Answer:Divide $x^3 + 5x^2 - 29x - 105$ by $x + 3$. - Junior Cycle Mathematics - Question b

Long Division Setup

First Division

Subtracting the First Product

Next Terms Division

Multiply and Subtract Again

Final Division

Final Result

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