Quadratic expressions

A quadratic expression must have highest power 2, so the coefficient \(a\) of \(x^2\) cannot be zero. If \(a=0\), the expression becomes at most linear. Exam tip: first check the coefficient of the highest-power term.

A constant term is a term whose value does not contain the variable. In a polynomial, the coefficient of the highest power is different from the constant term, and a term involving the variable is never constant. The sign belongs to the term and must be included when identifying it.

In \(2y^2+9y-11\), the term \(2y^2\) contains \(y^2\), so 2 is the quadratic coefficient. The term \(9y\) contains \(y\), so 9 is the linear coefficient. The term \(-11\) contains no variable, making it the constant term. Therefore option C is correct; writing 11 without its negative sign would be incomplete.

A perfect square is an expression formed by multiplying a binomial by itself. The identity \((a+b)^2=a^2+2ab+b^2\) helps us recognize it. Here the first term is \(x^2\), so one part is \(x\); the constant term is 9, whose square root is 3. Therefore the matching form must involve \(x+3\), not \(x-3\) or a larger constant.

Expanding \(\left(x+3\right)^2\) gives \(x^2+2\cdot x\cdot3+3^2=x^2+6x+9\). Both the middle term and the final term match exactly, so option A is correct. Option B would produce \(x^2-6x+9\), while options C and D give incorrect middle and constant terms. Thus the given expression is \(\left(x+3\right)^2\).

The middle term (-10a=-2\cdot a\cdot5), so the sign is negative. In exams the sign of the middle term is decisive.

To factor \(x^2+8x+15\), find two numbers whose product is the constant term \(15\) and whose sum is the middle coefficient \(8\). The positive factor pairs of 15 are 1 and 15, or 3 and 5. Only 3 and 5 have sum 8, so they give the required two linear factors.

Thus the factorised form is \( (x+3)(x+5) \). Expanding it gives \(x^2+5x+3x+15=x^2+8x+15\), which confirms choice A. The pair 1 and 15 has the wrong sum, while negative factors would produce a negative middle term. The repeated factors \(x+4\) would give constant term 16, not 15.

(-3+(-4)=-7) and ((-3)(-4)=12). In exams use two negative signs when the constant is positive and the middle term is negative.

The direct answer is option A: \((x+5)(x-4)\). For an expression of the form \(x^2+bx+c\), find two numbers whose product is \(c=-20\) and whose sum is \(b=1\). The numbers are \(5\) and \(-4\), because \(5\times(-4)=-20\) and \(5+(-4)=1\). Therefore \(x^2+x-20=(x+5)(x-4)\). Option A is correct. Option B uses \(-5\) and \(4\); their product is also \(-20\), but their sum is \(-1\), giving \(x^2-x-20\), not the given expression. Option C has product \(-20\), but \(2+(-10)=-8\), so its middle term is wrong. Option D has product \(-20\), but \(1+(-20)=-19\), so it also gives the wrong middle term. When the constant is negative, the two numbers have opposite signs; choose their signs so their sum has the required sign.

In ((3x+4)(x+2)), the cross terms are (6x) and (4x). In exams their sum should be the middle term.

Here, \(4x^2=(2x)^2\) and \(9=3^2\). Therefore, \(4x^2-9=(2x)^2-3^2\) is a difference of squares. Applying \(a^2-b^2=(a-b)(a+b)\) gives \((2x-3)(2x+3)\). Option \((2x-3)^2\) expands with a middle term \(-12x\), so it is not correct. Exam tip: for a difference of squares, take the square roots of both terms and write their difference and sum as factors.

The direct answer is B: \((3y-4)^2\). Use \((a-b)^2=a^2-2ab+b^2\). The first term \(9y^2\) is \((3y)^2\), and the last term 16 is \(4^2\). Their middle term must be \(-2\cdot3y\cdot4=-24y\), exactly as given. Thus \(9y^2-24y+16=(3y-4)^2\). Option A, \((3y+4)^2\), gives a positive middle term \(+24y\), so the sign is wrong. Option B gives the correct first term, constant term, and negative middle term. Option C, \((9y-4)^2\), has first term \(81y^2\), so it uses the wrong square root. Option D, \((3y-8)(3y-2)\), gives middle terms \(-6y-24y=-30y\) and constant 16, so its middle coefficient is wrong. For such questions, compare all three parts: first square, middle term, and last square. This prevents choosing an answer that matches only two parts.

The direct answer is option A, 0. Substitute x=2 carefully: x²−5x+6=2²−5(2)+6. First calculate the power, 2²=4. Then calculate the multiplication, 5(2)=10. The expression becomes 4−10+6. Now 4−10=−6, and −6+6=0. Option A is correct. Option B, 2, may result from an arithmetic or substitution mistake. Option C, 4, is only the value of x² and ignores the other terms. Option D, 6, is only the constant term and ignores the variable terms. The order of operations is important: powers first, multiplication next, and addition or subtraction from left to right. A useful check is that x²−5x+6=(x−2)(x−3), so x=2 makes one factor zero and the whole expression zero.

In option A, \((x+1)(x-1)+1=x^2-1+1=x^2\), whose highest power is 2, so it is quadratic. In option B, all variable terms cancel and only 1 remains. Exam tip: check the highest power after simplifying.

(16=4^2), and in a perfect square the middle term is (2\cdot x\cdot4=8x). In exams double the root of the constant term.

(-14x=-2\cdot x\cdot7), so (c=7^2=49). In exams take half of the middle coefficient and square it.

((x+3)(x+7)=x^2+10x+21), so (p=10). In exams the middle term comes from the sum of the constants.

In 3x² + 5x + 8, the coefficients of x² and x are 3 and 5. Their sum is 3 + 5 = 8, which equals the constant term. In option B, 2 + 7 = 9, not 10. Exam tip: include signs while adding coefficients.

Use the distributive property: \((x-4)(x+2)=x(x+2)-4(x+2)=x^2+2x-4x-8=x^2-2x-8\). Therefore, the correct expression is \(x^2-2x-8\). In \(x^2+6x-8\), the negative sign of \(-4x\) has been handled incorrectly. Exam tip: write all four products first, then combine like terms.

The direct answer is option A: \(2x^2+7x-15\). Use distribution and write all four products: \(2x\cdot x=2x^2\), \(2x\cdot5=10x\), \((-3)\cdot x=-3x\), and \((-3)\cdot5=-15\). Combining the middle terms gives \(10x-3x=7x\), so the expansion is \(2x^2+7x-15\). Option A is correct. Option B includes \(10x\) but forgets the other cross product \(-3x\). Option C incorrectly makes the middle term \(-13x\), usually from treating \(-3x\) and \(10x\) as both negative. Option D changes the constant product to \(+15\), although a negative number times a positive number is negative. Memory cue: in a binomial product, check two cross terms and the final sign separately.

Taking common (5) gives (5(x^2-4x+4)). In exams first take the common factor and then apply identities.

First take out (3), and (x^2-9) is a difference of squares. In exams factorise the inside expression too.

(4x^2=(2x)^2) and (12x=2\cdot2x\cdot3), so (k=9). In exams identify the second part using (2ab).

(9x^2=(3x)^2), (25=5^2), so the middle term is (-2\cdot3x\cdot5=-30x). In exams follow the given sign.

Here, \(9a^2=(3a)^2\) and \(16b^2=(4b)^2\). The middle term is \(-2\times3a\times4b=-24ab\), so the expression is \((3a-4b)^2\). Exam tip: always verify the middle coefficient.

A perfect-square trinomial has the form \(a^2\pm2ab+b^2\). Here, \(9p^2=(3p)^2\) and \(16q^2=(4q)^2\), while \(-2(3p)(4q)=-24pq\). Therefore, \(9p^2-24pq+16q^2=(3p-4q)^2\), so option A is correct. In option B, the last term is \(15q^2\) instead of \(16q^2\), so it is not a perfect square. Exam tip: take the square roots of the first and last terms, then check whether the middle term equals \(\pm2ab\).

Actually ((3p-q)(2p+q)) gives (6p^2+pq-q^2). Check the cross terms carefully in exams.