Question

1. solve the trinomial by factoring: $x^2 - 8x + 12 = 0$
2. factor $x^2 - 4x - 5$
3. factor $x^2 - x - 2$
4. factor $x^2 + 2x - 80$
5. factor $x^2 + 9x - 10$

Explanation:

Question 6:

Step1: Find two numbers

We need two numbers that multiply to \(12\) (the constant term) and add up to \(-8\) (the coefficient of \(x\)). The numbers are \(-6\) and \(-2\) since \((-6)\times(-2) = 12\) and \((-6)+(-2)=-8\).

Step2: Factor the trinomial

Rewrite the middle term using these two numbers: \(x^{2}-6x - 2x+12 = 0\). Then factor by grouping: \(x(x - 6)-2(x - 6)=0\), which gives \((x - 6)(x - 2)=0\).

Step3: Solve for \(x\)

Set each factor equal to zero: \(x - 6 = 0\) or \(x - 2 = 0\). So \(x = 6\) or \(x = 2\).

Step1: Find two numbers

We need two numbers that multiply to \(-5\) (the constant term) and add up to \(-4\) (the coefficient of \(x\)). The numbers are \(-5\) and \(1\) since \((-5)\times1=-5\) and \((-5)+1=-4\).

Step2: Factor the trinomial

Rewrite the middle term using these two numbers: \(x^{2}-5x+x - 5\). Then factor by grouping: \(x(x - 5)+1(x - 5)=(x - 5)(x + 1)\).

Step1: Find two numbers

We need two numbers that multiply to \(-2\) (the constant term) and add up to \(-1\) (the coefficient of \(x\)). The numbers are \(-2\) and \(1\) since \((-2)\times1=-2\) and \((-2)+1=-1\).

Step2: Factor the trinomial

Rewrite the middle term using these two numbers: \(x^{2}-2x+x - 2\). Then factor by grouping: \(x(x - 2)+1(x - 2)=(x - 2)(x + 1)\).

Answer:

\(x = 2\) or \(x = 6\)

Question 7: