If \(p(x)=x^2+kx+16\) is a perfect-square polynomial and \(k<0\), what is the value of \(k\)?
Answer and explanation
Correct answer: -8
A perfect-square polynomial with leading coefficient 1 has the form \((x+a)^2=x^2+2ax+a^2\). Since the constant term is \(16=4^2\), we get \(p(x)=(x-4)^2=x^2-8x+16\), so \(k=-8\). The condition \(k<0\) rules out \(k=8\). Exam tip: For \(x^2+bx+c^2\) to be a perfect square, the middle coefficient must be \(\pm 2c\).
Frequently asked questions
What is the correct answer to this question?
-8
Why is this the correct answer?
A perfect-square polynomial with leading coefficient 1 has the form \((x+a)^2=x^2+2ax+a^2\). Since the constant term is \(16=4^2\), we get \(p(x)=(x-4)^2=x^2-8x+16\), so \(k=-8\). The condition \(k<0\) rules out \(k=8\). Exam tip: For \(x^2+bx+c^2\) to be a perfect square, the middle coefficient must be \(\pm 2c\).
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Polynomials. Topic: Polynomials in one variable.
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