Answer :
Certainly! Let's factor the expression [tex]\(4x^2 + 28x^3\)[/tex].
### Step-by-Step Solution
1. Identify the Greatest Common Factor (GCF):
- Look at the coefficients and variables in the terms [tex]\(4x^2\)[/tex] and [tex]\(28x^3\)[/tex].
- The numeric coefficients are 4 and 28. The GCF of 4 and 28 is 4.
- For the variables, we have [tex]\(x^2\)[/tex] in the first term and [tex]\(x^3\)[/tex] in the second term. The GCF for the variables is [tex]\(x^2\)[/tex] because it's the highest power of [tex]\(x\)[/tex] that divides both.
2. Extract the GCF:
- Factor out [tex]\(4x^2\)[/tex] from each term in the expression.
- To factor each term:
- From [tex]\(4x^2\)[/tex]: [tex]\(4x^2 \div 4x^2 = 1\)[/tex]
- From [tex]\(28x^3\)[/tex]: [tex]\(28x^3 \div 4x^2 = 7x\)[/tex]
3. Write the Factored Expression:
- After factoring out the GCF, the expression becomes:
[tex]\[
4x^2(1 + 7x)
\][/tex]
4. Conclusion:
- The expression [tex]\(4x^2 + 28x^3\)[/tex] factors to [tex]\(4x^2(1 + 7x)\)[/tex].
I hope this explanation helps! Feel free to ask if you have any more questions.
### Step-by-Step Solution
1. Identify the Greatest Common Factor (GCF):
- Look at the coefficients and variables in the terms [tex]\(4x^2\)[/tex] and [tex]\(28x^3\)[/tex].
- The numeric coefficients are 4 and 28. The GCF of 4 and 28 is 4.
- For the variables, we have [tex]\(x^2\)[/tex] in the first term and [tex]\(x^3\)[/tex] in the second term. The GCF for the variables is [tex]\(x^2\)[/tex] because it's the highest power of [tex]\(x\)[/tex] that divides both.
2. Extract the GCF:
- Factor out [tex]\(4x^2\)[/tex] from each term in the expression.
- To factor each term:
- From [tex]\(4x^2\)[/tex]: [tex]\(4x^2 \div 4x^2 = 1\)[/tex]
- From [tex]\(28x^3\)[/tex]: [tex]\(28x^3 \div 4x^2 = 7x\)[/tex]
3. Write the Factored Expression:
- After factoring out the GCF, the expression becomes:
[tex]\[
4x^2(1 + 7x)
\][/tex]
4. Conclusion:
- The expression [tex]\(4x^2 + 28x^3\)[/tex] factors to [tex]\(4x^2(1 + 7x)\)[/tex].
I hope this explanation helps! Feel free to ask if you have any more questions.