Quadratic expressions

For \(x^2-5x+6\), we need two integers whose product is 6 and whose sum is \(-5\). They are \(-2\) and \(-3\), so it is \((x-2)(x-3)\). Exam tip: check both product and sum.

A quadratic polynomial must have a non-zero coefficient of \(x^2\). In option C, \(a^2+1>0\) for every real \(a\). In B, taking \(a=0\) removes the quadratic term. Exam tip: always check whether the leading coefficient can become zero.

((x+2)(x+8)=x^2+10x+16) and ((x+5)^2=x^2+10x+25). Exam tip: cancel common terms and compare constants.

In the expansion, the middle term is (4x+6x=10x). Exam tip: add the constants of the binomials.

Using \(a^2-2ab+b^2=(a-b)^2\), take \(a=4p\) and \(b=3q\). The middle term is \(-2(4p)(3q)=-24pq\), so A is \((4p-3q)^2\). Exam tip: verify the middle term after checking the square terms.

The first constant term is (-25), and the second is (25), so the difference is (-50). Exam tip: do not mix difference of squares with perfect square.

\(x^2+2ax+a^2=(x+a)^2\), so it is a perfect square for every real \(a\). In option B, the middle term is \(ax\), not \(2ax\). Exam tip: the middle term of \((x+a)^2\) is always \(2ax\).

Here \(9p^2=(3p)^2\) and \(4q^2=(2q)^2\), while the middle term is \(2(3p)(2q)=12pq\). Thus A is \((3p+2q)^2\). Option B has 10pq instead. Exam tip: compare with \(a^2+2ab+b^2\).

\(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=(x+5)^2\), so A is a perfect square. In B, the constant term would need to be \(25\). Exam tip: match the middle term with \(2ab\).

The value is ((-1)^2+7^2=1+49=50), not (58). Exam tip: recheck calculation against options.

\(x^2+2x+2=(x+1)^2+1\). Since \((x+1)^2\geq 0\), its minimum value is \(1\), so the expression is strictly positive for every real \(x\). The closest distractor, option B, is \((x-1)^2\), which equals \(0\) at \(x=1\); hence it is not strictly positive. Exam tip: complete the square to find the minimum value of a quadratic.

In (-2x^2+7x+3), the coefficient of (x^2) is (-2). Exam tip: changing the order of terms does not change the coefficient.

\(x^2+4x+4=(x+2)^2=(x+2)(x+2)\), so it is a product of two linear factors. For option B, the discriminant is \(b^2-4ac=16-20=-4\), so it has no real linear factors. Exam tip: spot perfect-square trinomials.

At (x=0), the value is (1), and at (x=2), the value is (1), so the sum is (2). Exam tip: find both values separately and add.

(x^2+2x+5=(x+1)^2+4), so (a+b=5). Exam tip: complete the square and add the remaining constant.

Since (36=6^2), the negative middle term is (-2\cdot6\cdot x=-12x). Exam tip: choose the sign according to the condition.

Using ((a+b)^2-(a-b)^2=4ab) with (a=2x) and (b=5) gives (40x). Exam tip: apply the identity directly.

The difference is (3x^2+2x-5-(x^2-4x+7)=2x^2+6x-12). Exam tip: change signs inside the bracket while subtracting.

Squaring gives (x^2+4+\frac{4}{x^2}=36), so the value is (32). Exam tip: subtract the (2ab) term (4).

((x-1)(x-9)=x^2-10x+9) and ((x-5)^2=x^2-10x+25). Exam tip: remove common terms and compare constants.