Quadratics by factoring (intro) - Questions

Answers

1.

\( \text { lesser x} = 1\)

\( \text { greater x} = 7\)

To factor \(x^{2}-8 x+7 \text { as }(x+a)(x+b)\), we need to find numbers \(a\) and \(b\) such that \(a+b=−8\) and \(ab = 7\).

\(a=−7\) and \(b=-1\) satisfy both conditions, so our equation can be re-written:

\((x−7)(x−1)=0\)

According to the zero-product property, we know that

\(x−7=0\) or \(x - 1 = 0\)

which means

\(x=7x\) or \(x=1\)

In conclusion,

\( \text { lesser x} = 1\)

\( \text { greater x} = 7\)


2. \(x = -8\)

Both \(x^{2}\) and \(64\) are perfect squares, since \(x^{2}=(x)^{2}\) and \(64=(8)^{2}\).

Additionally, \(16x\) is twice the product of the roots of \(x^{2}\) and \(64\), since \(16 x=2(x)(8)\).

\(x^{2}+16 x+64=(x)^{2}+2(x)(8)+(8)^{2}\)

So we can use the square of a sum pattern to factor:

\(a^{2}+2(a)(b)+b^{2}=(a+b)^{2}\)

In this case, \(a = x\) and \(b = 8\):

\((x)^{2}+2(x)(8)+(8)^{2}=(x+8)^{2}\)

So our equation can be re-written:

\((x+8)^{2}=0\)

The only possible solution is when \(x+8 = 0\), which is

\(x = -8\)


3.

\( \text { lesser x} = -9\)

\( \text { greater x} = -3\)

To factor \(x^{2}+12 x+27\) as \((x+a)(x+b)\), we need to find numbers \(a\) and \(b\) such that

\(a+b=12\) and \(a b=27\).

\(a = 3\) and \(b = 9\) satisfy both conditions, so our equation can be re-written:

\((x+3)(x+9)=0\)

According to the zero-product property, we know that

\(x+3=0\) or \(x+9 = 0\)

which means

\(x=−3\) or \(x=-9\)

In conclusion,

\( \text { lesser x} = -9\)

\( \text { greater x} = -3\)


4. \(x = -5\)

Both \(x^{2}\) and \(25\) are perfect squares, since \(x^{2}=(x)^{2} \text { and } 25=(5)^{2}\).

Additionally, \(10x\) is twice the product of the roots of \(x^{2} \text { and } 25, \text { since } 10 x=2(x)(5)\).

\(x^{2}+10 x+25=(x)^{2}+2(x)(5)+(5)^{2}\)

So we can use the square of a sum pattern to factor:

\(a^{2}+2(a)(b)+b^{2}=(a+b)^{2}\)

In this case, \(a = x\) and \(b = 5\):

\((x)^{2}+2(x)(5)+(5)^{2}=(x+5)^{2}\)

So our equation can be re-written:

\((x+5)^{2}=0\)

The only possible solution is when \(x + 5 = 0\), which is

\(x = -5\)