Question

question 1-13
which of the following are factors of $a^2 - 60a + 900$?
i. $a - 20$
ii. $a + 30$
iii. $a - 30$
iv. $a + 20$
\bigcirc ii only
\bigcirc iii only
\bigcirc i and iv only
\bigcirc ii and iv only

Explanation:

Step1: Recall the perfect square trinomial formula

The perfect square trinomial formula is \(x^2 - 2xy + y^2=(x - y)^2\). For the quadratic expression \(a^2-60a + 900\), we can compare it with the form \(x^2-2xy + y^2\). Here, \(x = a\), and we need to find \(y\) such that \(2xy=60a\) and \(y^2 = 900\).
From \(y^2=900\), we get \(y = 30\) (since \((30)^2=900\)). And \(2xy=2\times a\times30 = 60a\), which matches the middle term. So, \(a^2-60a + 900=(a - 30)^2=(a - 30)(a - 30)\).

Step2: Check each factor

• For I. \(a - 20\): If we divide \(a^2-60a + 900\) by \(a - 20\), we can use polynomial long - division or the factor theorem. Let \(f(a)=a^2-60a + 900\). Then \(f(20)=20^2-60\times20 + 900=400-1200 + 900 = 100

eq0\), so \(a - 20\) is not a factor.

• For II. \(a + 30\): Let \(f(a)=a^2-60a + 900\). Then \(f(-30)=(-30)^2-60\times(-30)+900=900 + 1800+900=3600

eq0\), so \(a + 30\) is not a factor.

• For III. \(a - 30\): From the factoring in Step 1, we know that \(a^2-60a + 900=(a - 30)(a - 30)\), so \(a - 30\) is a factor.
• For IV. \(a + 20\): Let \(f(a)=a^2-60a + 900\). Then \(f(-20)=(-20)^2-60\times(-20)+900=400 + 1200+900=2500

eq0\), so \(a + 20\) is not a factor.

Answer:

III only