Question

j) $(15x^3y^2 - x^2y + 5xy) div 5xy$

1. find a simplified algebraic expression for the perimeter of each shape below: (2 marks each)

a)

triangle with sides $x + 14$, $x + 14$, $2x + 36$

b)

rectangle with length $2x^2 - x + 3$ and width $x^2 - 3x + 4$

Explanation:

Part (i): Simplify \((15x^{3}y^{2}-x^{2}y + 5xy)\div5xy\)

Step 1: Divide each term by \(5xy\)

We use the rule of dividing a polynomial by a monomial, which is to divide each term of the polynomial by the monomial. So we have:
\(\frac{15x^{3}y^{2}}{5xy}-\frac{x^{2}y}{5xy}+\frac{5xy}{5xy}\)

Step 2: Simplify each term

• For the first term \(\frac{15x^{3}y^{2}}{5xy}\), we divide the coefficients and subtract the exponents of like bases. \(15\div5 = 3\), \(x^{3 - 1}=x^{2}\), \(y^{2-1}=y\), so it simplifies to \(3x^{2}y\).
• For the second term \(\frac{x^{2}y}{5xy}\), \(x^{2-1}=x\), \(y^{1 - 1}=y^{0}=1\), and the coefficient is \(\frac{1}{5}\), so it simplifies to \(\frac{x}{5}\).
• For the third term \(\frac{5xy}{5xy}\), both the numerator and denominator are the same, so it simplifies to \(1\).

Putting it all together, we get \(3x^{2}y-\frac{x}{5}+1\)

Step 1: Recall the formula for the perimeter of a triangle

The perimeter \(P\) of a triangle is the sum of the lengths of its three sides. The sides of the triangle are \(x + 14\), \(x+14\) and \(2x + 36\).

Step 2: Sum the side lengths

\(P=(x + 14)+(x + 14)+(2x+36)\)

Step 3: Combine like terms

First, remove the parentheses: \(x+14+x + 14+2x+36\)

Then, combine the \(x\) terms: \(x+x + 2x=4x\)

And combine the constant terms: \(14 + 14+36=64\)

So the perimeter is \(4x + 64\)

Step 1: Recall the formula for the perimeter of a rectangle

The perimeter \(P\) of a rectangle is given by \(P = 2\times(\text{length}+\text{width})\). The length is \(2x^{2}-x + 3\) and the width is \(x^{2}-3x + 4\).

Step 2: Substitute into the formula

\(P=2\times[(2x^{2}-x + 3)+(x^{2}-3x + 4)]\)

Step 3: Simplify the expression inside the parentheses

Combine like terms: \((2x^{2}+x^{2})+(-x-3x)+(3 + 4)=3x^{2}-4x + 7\)

Step 4: Multiply by 2

\(P = 2\times(3x^{2}-4x + 7)=6x^{2}-8x + 14\)

Answer:

\(3x^{2}y-\frac{x}{5}+1\)

Part 3(a): Perimeter of Triangle \(ABC\)