Question

drag the tiles to the boxes to form correct pairs. not all tiles will be used.
match the trinomials with their factors.
$a^2 + a - 20$ $a^2 - 9a + 20$ $a^2 - 8a - 20$ $a^2 - 12a + 20$ $a^2 - 19a - 20$
factors trinomials
$(a - 4)(a - 5)$ $longrightarrow$
$(a - 10)(a - 2)$ $longrightarrow$
$(a - 4)(a + 5)$ $longrightarrow$
$(a - 10)(a + 2)$ $longrightarrow$

Explanation:

Step1: Expand \((a - 4)(a - 5)\)

Using the distributive property \((x + m)(x + n)=x^2+(m + n)x+mn\), here \(x = a\), \(m=-4\), \(n = - 5\). So \((a - 4)(a - 5)=a^{2}+(-4-5)a+(-4)\times(-5)=a^{2}-9a + 20\).

Step2: Expand \((a - 10)(a - 2)\)

Using the distributive property, \(x=a\), \(m = - 10\), \(n=-2\). So \((a - 10)(a - 2)=a^{2}+(-10-2)a+(-10)\times(-2)=a^{2}-12a + 20\).

Step3: Expand \((a - 4)(a + 5)\)

Using the distributive property, \(x = a\), \(m=-4\), \(n = 5\). So \((a - 4)(a + 5)=a^{2}+(-4 + 5)a+(-4)\times5=a^{2}+a-20\).

Step4: Expand \((a - 10)(a + 2)\)

Using the distributive property, \(x=a\), \(m=-10\), \(n = 2\). So \((a - 10)(a + 2)=a^{2}+(-10 + 2)a+(-10)\times2=a^{2}-8a-20\).

Answer:

\((a - 4)(a - 5)\) pairs with \(a^{2}-9a + 20\)

\((a - 10)(a - 2)\) pairs with \(a^{2}-12a + 20\)

\((a - 4)(a + 5)\) pairs with \(a^{2}+a-20\)

\((a - 10)(a + 2)\) pairs with \(a^{2}-8a-20\)