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^210s2+28st−6t2

Step 1: Look for a greatest common factor (GCF)

The terms 10s210s^210s2, $28st$, and −6t2-6t^2−6t2 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 $10$, and the constant is $-6$. Multiply them:

$$10\times -6=-60$$

Step 3: Find two numbers that multiply to $-60$ and add to $28$

The two numbers are $30$ and $-2$, because:

$$30\times -2=-60\text{and}30+(-2)=28$$

Step 4: Rewrite the middle term using these numbers

10s2+30st−2st−6t210s^2 + 30st – 2st – 6t^210s2+30st−2st−6t2

Step 5: Group terms and factor each group

Group the terms:

(10s2+30st)+(−2st−6t2)(10s^2 + 30st) + (-2st – 6t^2)(10s2+30st)+(−2st−6t2)

Factor out the GCF from each group:

$$10s(s+3t)-2t(s+3t)$$

Step 6: Factor out the common binomial

$$(10s-2t)(s+3t)$$

Final Answer:

10s2+28st−6t2=(10s−2t)(s+3t)10s^2 + 28st – 6t^2 = (10s – 2t)(s + 3t)10s2+28st−6t2=(10s−2t)(s+3t)

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• 6z3+3z2+2z+16z^3 + 3z^2 + 2z + 16z3+3z2+2z+1

Step 1: Group terms for factoring by grouping

Group the terms:

(6z3+3z2)+(2z+1)(6z^3 + 3z^2) + (2z + 1)(6z3+3z2)+(2z+1)

Step 2: Factor each group

From the first group, 6z3+3z26z^3 + 3z^26z3+3z2, factor out 3z23z^23z2:

3z2(2z+1)3z^2(2z + 1)3z2(2z+1)

From the second group, $2z+1$, factor out $1$ (since there’s no other common factor):

$$1(2z+1)$$

Step 3: Combine the factored terms

(3z2+1)(2z+1)(3z^2 + 1)(2z + 1)(3z2+1)(2z+1)

Final Answer:

6z3+3z2+2z+1=(3z2+1)(2z+1)6z^3 + 3z^2 + 2z + 1 = (3z^2 + 1)(2z + 1)6z3+3z2+2z+1=(3z2+1)(2z+1)

 

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