Q. Simplify the expression: \(7x^{2} + 3 – 5(x^{2} – 4)\)
Answer
We interpret the expression as \(7x^2 + 3 – 5(x^2 – 4)\). Distribute:
\[
7x^2 + 3 – 5(x^2 – 4) = 7x^2 + 3 – 5x^2 + 20
\]
\[
= 2x^2 + 23
\]
Final result: \(\boxed{2x^2 + 23}\).
Detailed Explanation
Problem: Simplify: \(7x^{2} + 3 – 5\bigl(x^{2} – 4\bigr)\)
Step 1 – Apply the distributive property
Multiply \(-5\) by each term in the parentheses:
\[
-5\cdot\bigl(x^{2}-4\bigr) = -5x^{2} + 20
\]
Substitute this into the original expression:
\[
7x^{2} + 3 – 5\bigl(x^{2}-4\bigr) = 7x^{2} + 3 – 5x^{2} + 20
\]
Step 2 – Group like terms
Group the terms with \(x^{2}\) and the constant terms:
\[
7x^{2} – 5x^{2} + 3 + 20
\]
Step 3 – Combine like terms
\[
7x^{2} – 5x^{2} = 2x^{2}, \qquad 3 + 20 = 23
\]
So the simplified expression is:
\[
2x^{2} + 23
\]
Answer
\[
7x^{2} + 3 – 5\bigl(x^{2}-4\bigr) = 2x^{2} + 23
\]
See full solution
FAQs
How do I simplify (7x^2+3-5(x^2-4))?
Distribute (-5): (-5x^2+20). Combine: (7x^2-5x^2=2x^2) and (3+20=23). Result: (2x^2+23).
Which property removes the parentheses?
The distributive property: (a(b+c)=ab+ac). Here (-5(x^2-4)=-5x^2+20).
How do I combine like terms?
Add coefficients of same powers: (7x^2-5x^2=2x^2). Combine constants: (3+20=23). So (2x^2+23).
Can (2x^2+23) be factored further?
Not over the integers. You can write (2x^2+23=2(x^2+11.5)), but there is no nontrivial integer-factorization.
What is the degree and leading coefficient?
Degree is 2 and the leading coefficient is 2 for (2x^2+23).
How can I check my simplification?
Substitute a value, e.g. (x=1): original gives (25); simplified gives (2(1)^2+23=25). They match.
What common sign mistake should I avoid?
Don’t forget the negative distributes: (-5(x^2-4)) becomes (-5x^2+20). A common error is writing (-5x^2-20).
What is the domain of the simplified expression?
All real numbers. Polynomials like (2x^2+23) are defined for every real (x).
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