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 formula for a perfect square trinomial is \(x^{2}-2xy + y^{2}=(x - y)^{2}\). In the given expression \(a^{2}-60a + 900\), we can compare it with the formula. Here, \(x=a\) and \(2xy = 60a\), so \(2y=60\) which gives \(y = 30\) (since \(y^{2}=900\) and \(\sqrt{900}=30\)). So, \(a^{2}-60a + 900=(a - 30)^{2}=(a - 30)(a - 30)\).

Step2: Check each factor

• For factor I: \(a - 20\). If we divide \((a - 30)^{2}=a^{2}-60a + 900\) by \(a - 20\), we can use polynomial long division or the factor theorem. The factor theorem states that if \(f(a)=(a - 20)\) is a factor, then \(f(20)=0\). 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 factor II: \(a + 30\). Using the factor theorem, if \(f(-30)=0\), then \(a + 30\) is a factor. \(f(-30)=(-30)^{2}-60\times(-30)+900=900 + 1800+900=3600

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

• For factor III: \(a - 30\). From the factoring above, we know that \((a - 30)\) is a factor of \((a - 30)^{2}\).
• For factor IV: \(a + 20\). Using the factor theorem, \(f(-20)=(-20)^{2}-60\times(-20)+900=400 + 1200+900=2500

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

Answer:

III only