Question

ratic *check for gcf
11x² - 8x - 3 = 0
p: s:
signs:
factor:
solve:

Explanation:

Step1: Find P and S

For the quadratic \(11x^2 - 8x - 3 = 0\), in the form \(ax^2+bx + c\), \(a = 11\), \(b=-8\), \(c = -3\).
\(P=a\times c=11\times(-3)= - 33\)
\(S = b=-8\)
Signs: One positive, one negative (since \(P=-33\) is negative, so factors have opposite signs; and \(S=-8\) is negative, so the larger factor in absolute value is negative).

Step2: Find factors of P that sum to S

We need two numbers that multiply to \(-33\) and add to \(-8\). The numbers are \(-11\) and \(3\) (since \(-11\times3=-33\) and \(-11 + 3=-8\)).

Step3: Split the middle term

Rewrite the quadratic as \(11x^2-11x + 3x-3 = 0\).

Step4: Factor by grouping

Group the first two and last two terms:
\((11x^2-11x)+(3x - 3)=0\)
Factor out GCF from each group:
\(11x(x - 1)+3(x - 1)=0\)
Factor out \((x - 1)\):
\((11x + 3)(x - 1)=0\)

Step5: Solve for x

Set each factor equal to zero:
\(11x+3 = 0\) or \(x - 1=0\)
For \(11x+3 = 0\), \(11x=-3\), so \(x=-\frac{3}{11}\)
For \(x - 1=0\), \(x = 1\)

Answer:

• P: \(-33\)
• S: \(-8\)
• Signs: One positive, one negative (larger absolute value negative)
• Factor: \((11x + 3)(x - 1)=0\)
• Solve: \(x = 1\) or \(x=-\frac{3}{11}\)