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write the perimeter of the triangle as a simplified polynomial. then fa…

Question

write the perimeter of the triangle as a simplified polynomial. then factor the polynomial. the perimeter of the triangle is (simplify your answer.)

Explanation:

Step1: Find the perimeter

The perimeter \(P\) of a triangle is the sum of its side lengths. So \(P=(5x^{2}+1)+(12x + 6)+(x^{2}+31x)\).
Combine like - terms:

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Step2: Factor the polynomial

For a quadratic polynomial \(ax^{2}+bx + c\) (\(a = 6\), \(b=43\), \(c = 7\)), we use the formula \(ax^{2}+bx + c=a(x - x_1)(x - x_2)\), where \(x_{1,2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(43)^{2}-4\times6\times7=1849 - 168 = 1681\).
Then \(x=\frac{-43\pm\sqrt{1681}}{12}=\frac{-43\pm41}{12}\).
We get \(x_1=\frac{-43 + 41}{12}=-\frac{1}{6}\) and \(x_2=-7\).
So \(6x^{2}+43x + 7=6(x+\frac{1}{6})(x + 7)=(6x + 1)(x+7)\)

Answer:

The perimeter of the triangle is \(6x^{2}+43x + 7\) and its factored form is \((6x + 1)(x + 7)\)