Question

which expression is prime?
a ( x^2 - 4x + 3 )
b ( x^2 - 6x + 9 )
c ( 3x^2 + 5x - 2 )
d ( 6x^2 - 7x + 10 )

Explanation:

Step1: Factor Option A

We factor \(x^{2}-4x + 3\). We need two numbers that multiply to \(3\) and add to \(- 4\). The numbers are \(-1\) and \(-3\). So \(x^{2}-4x + 3=(x - 1)(x - 3)\).

Step2: Factor Option B

For \(x^{2}-6x + 9\), this is a perfect square trinomial. Using the formula \((a - b)^{2}=a^{2}-2ab + b^{2}\), where \(a=x\) and \(b = 3\), we have \(x^{2}-6x + 9=(x - 3)^{2}\).

Step3: Factor Option C

For \(3x^{2}+5x - 2\), we use the AC method. \(A = 3\), \(B=5\), \(C=-2\). \(AC=3\times(-2)=-6\). We need two numbers that multiply to \(-6\) and add to \(5\). The numbers are \(6\) and \(-1\). So we rewrite the middle term: \(3x^{2}+6x-x - 2=3x(x + 2)-1(x + 2)=(3x - 1)(x + 2)\).

Step4: Try to Factor Option D

For \(6x^{2}-7x + 10\), we use the AC method. \(A = 6\), \(B=-7\), \(C = 10\). \(AC=6\times10 = 60\). We need two numbers that multiply to \(60\) and add to \(-7\). The factors of \(60\) are \(1\) & \(60\), \(2\) & \(30\), \(3\) & \(20\), \(4\) & \(15\), \(5\) & \(12\), \(6\) & \(10\). None of these pairs add up to \(-7\) (since both numbers would have to be negative to multiply to \(60\) and add to \(-7\), but even the sum of the negative factors will not be \(-7\)). So \(6x^{2}-7x + 10\) cannot be factored (it is prime).

Answer:

D. \(6x^{2}-7x + 10\)