Factoring Quadratic Expressions
Factoring Quadratic Expressions
(Factoring Quadratic Expressions)
Math question.
For your assigned student number, factor the expressions using any method. If your polynomial cannot be factored, then label it “prime.” Be sure to explain your steps!
- 10s2 + 28st – 6t2
- 6z3 + 3z2 + 2z + 1
Soln.
- 10s2+28st−6t210s^2 + 28st – 6t^2
Step 1: Look for a greatest common factor (GCF)
The terms 10s210s^2, 28st28st, and −6t2-6t^2 do not share a common factor other than 1, so we proceed without factoring out a GCF.
Step 2: Multiply the leading coefficient and the constant
The leading coefficient is 1010, and the constant is −6-6. Multiply them:
10×−6=−6010 \times -6 = -60
Step 3: Find two numbers that multiply to −60-60 and add to 2828
The two numbers are 3030 and −2-2, because:
30×−2=−60and30+(−2)=2830 \times -2 = -60 \quad \text{and} \quad 30 + (-2) = 28
Step 4: Rewrite the middle term using these numbers
10s2+30st−2st−6t210s^2 + 30st – 2st – 6t^2
Step 5: Group terms and factor each group
Group the terms:
(10s2+30st)+(−2st−6t2)(10s^2 + 30st) + (-2st – 6t^2)
Factor out the GCF from each group:
10s(s+3t)−2t(s+3t)10s(s + 3t) – 2t(s + 3t)
Step 6: Factor out the common binomial
(10s−2t)(s+3t)(10s – 2t)(s + 3t)
Final Answer:
10s2+28st−6t2=(10s−2t)(s+3t)10s^2 + 28st – 6t^2 = (10s – 2t)(s + 3t)
- 6z3+3z2+2z+16z^3 + 3z^2 + 2z + 1
Step 1: Group terms for factoring by grouping
Group the terms:
(6z3+3z2)+(2z+1)(6z^3 + 3z^2) + (2z + 1)
Step 2: Factor each group
From the first group, 6z3+3z26z^3 + 3z^2, factor out 3z23z^2:
3z2(2z+1)3z^2(2z + 1)
From the second group, 2z+12z + 1, factor out 11 (since there’s no other common factor):
1(2z+1)1(2z + 1)
Step 3: Combine the factored terms
(3z2+1)(2z+1)(3z^2 + 1)(2z + 1)
Final Answer:
6z3+3z2+2z+1=(3z2+1)(2z+1)6z^3 + 3z^2 + 2z + 1 = (3z^2 + 1)(2z + 1)
