Question

1. the base of a triangle is $\frac{2x^{2}+5x - 3}{x^{2}-x - 12}$ and the height is $\frac{x^{2}-x - 2}{2x^{2}-5x + 2}$

Explanation:

1. First, factor the polynomials:

• Factor the numerator and denominator of the base - expression:
• For \(2x^{2}+5x - 3\), we have \(2x^{2}+5x - 3=2x^{2}+6x - x - 3 = 2x(x + 3)-(x + 3)=(2x - 1)(x + 3)\).
• For \(x^{2}-x - 12\), we have \(x^{2}-x - 12=x^{2}-4x+3x - 12=x(x - 4)+3(x - 4)=(x - 4)(x + 3)\).
• So, the base \(b=\frac{2x^{2}+5x - 3}{x^{2}-x - 12}=\frac{(2x - 1)(x + 3)}{(x - 4)(x + 3)}=\frac{2x - 1}{x - 4}\) (assuming \(x

eq - 3\)).

• Factor the numerator and denominator of the height - expression:
• For \(x^{2}-x - 2\), we have \(x^{2}-x - 2=x^{2}-2x+x - 2=x(x - 2)+(x - 2)=(x - 2)(x+1)\).
• For \(2x^{2}-5x + 2\), we have \(2x^{2}-5x + 2=2x^{2}-4x - x + 2=2x(x - 2)-(x - 2)=(2x - 1)(x - 2)\).
• So, the height \(h=\frac{x^{2}-x - 2}{2x^{2}-5x + 2}=\frac{(x - 2)(x + 1)}{(2x - 1)(x - 2)}=\frac{x + 1}{2x - 1}\) (assuming \(x

eq2\)).

1. Then, use the formula for the area of a triangle \(A=\frac{1}{2}bh\):

• Substitute \(b = \frac{2x - 1}{x - 4}\) and \(h=\frac{x + 1}{2x - 1}\) into the area formula.
• \(A=\frac{1}{2}\times\frac{2x - 1}{x - 4}\times\frac{x + 1}{2x - 1}\).
• Cancel out the common factor \((2x - 1)\):
• \(A=\frac{1}{2}\times\frac{x + 1}{x - 4}=\frac{x + 1}{2(x - 4)}\), where \(x

eq - 3,x
eq2,x
eq\frac{1}{2}\).

Step1: Factor polynomials in base

Factor \(2x^{2}+5x - 3=(2x - 1)(x + 3)\) and \(x^{2}-x - 12=(x - 4)(x + 3)\), then simplify base.

Step2: Factor polynomials in height

Factor \(x^{2}-x - 2=(x - 2)(x + 1)\) and \(2x^{2}-5x + 2=(2x - 1)(x - 2)\), then simplify height.

Step3: Calculate area

Use \(A=\frac{1}{2}bh\), cancel common factor \((2x - 1)\).

Answer:

\(\frac{x + 1}{2(x - 4)}\)