Quadratic expressions

For \(x^2-5x+6\), \(a=1, b=-5, c=6\), so \(b^2-4ac=25-24=1>0\). Hence it has two distinct real zeroes. In option B, the discriminant is 0, giving equal zeroes. Exam tip: check \(D>0\) for distinct real zeroes.

\(x^2-6x+9=(x-3)^2\). The square of every real number is zero or positive, so this trinomial can never be negative. Option B equals \((x-3)^2-1\), which can be negative. Exam tip: look for a perfect-square identity.

Since (x^2+x-12=(x+4)(x-3)), the values are (-4) and (3). Exam tip: do not get confused by the order in options.

The direct answer is option B: \(4x\). Interpret “difference” as the first expression minus the second: \((x^2+2x+1)-(x^2-2x+1)\). Remove brackets carefully: \(x^2+2x+1-x^2+2x-1\). The like terms cancel: \(x^2-x^2=0\), \(1-1=0\), and \(2x+2x=4x\). Thus the difference is \(4x\). Option A, \(2x\), misses one of the two linear terms. Option B is correct because both linear contributions combine to \(4x\). Option C, \(2\), is a constant and ignores the variable terms. Option D, \(0\), would result only if the two expressions were identical, but their middle terms have opposite signs. A common mistake is to subtract the second bracket without changing the signs inside it; every sign changes when a minus sign precedes a bracket.

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.

When two linear factors have the form \((x+r)(x+s)\), their product is \(x^2+(r+s)x+rs\). Here the stated factors are \((x+4)(x+6)\). The constant terms multiply to \(4\times6=24\), matching the last term, and the middle coefficient is the sum \(4+6=10\). Therefore the value of \(b\) is 10, which is option D.

Expanding makes the conclusion clear: \((x+4)(x+6)=x^2+6x+4x+24=x^2+10x+24\). Comparing this with \(x^2+bx+24\) gives \(b=10\). Option A is the constant term rather than the middle coefficient, while options B and C are only one of the two contributing numbers. The important method is to add the constants in the two binomials to find the coefficient of \(x\).

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.

To find the difference, subtract the entire second polynomial from the first one. The subtraction sign before the bracket changes every sign inside that bracket: x² becomes -x², -4x becomes +4x, and +7 becomes -7. Combining like terms gives 3x² - x² = 2x², 2x + 4x = 6x, and -5 - 7 = -12. The difference is therefore 2x² + 6x - 12.

The coefficient of x is the number multiplying x, so it is 6. Hence choice A is correct. Choosing 2 would mean forgetting that subtracting -4x creates +4x; choosing 4 would retain only the second polynomial’s changed contribution and ignore the original 2x term.

The direct answer is C: \(32\). Start with \(x+\frac{2}{x}=6\), where \(x\neq0\). Square both sides: \(\left(x+\frac{2}{x}\right)^2=6^2\). Expanding gives \(x^2+2\cdot x\cdot\frac{2}{x}+\frac{4}{x^2}=36\). The middle term simplifies to 4, so \(x^2+4+\frac{4}{x^2}=36\). Subtract 4 from both sides: \(x^2+\frac{4}{x^2}=32\). Option C is correct. Option A, 30, subtracts the wrong amount. Option B, 34, adds 2 instead of removing the middle term 4. Option D, 36, is the squared right-hand side before removing the extra middle term. The important point is that squaring creates a middle term, which must be handled. Memory cue: after squaring \(a+b\), remove \(2ab\) to obtain \(a^2+b^2\).

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