NCERT Exemplar Class 10 Maths Chapter 2 Polynomials Solutions

Correct option: (A) 43.

Concept used. A number is a zero of a polynomial when putting it in for x makes the polynomial equal to 0. So substitute x=-3 and solve the resulting equation for k.

1. Put x=-3 into (k-1)x²+kx+1 and set it equal to 0: [] (k-1)(-3)²+k(-3)+1=0 [] (k-1)(9)-3k+1=0 [] 9k-9-3k+1=0 [] 6k-8=0
2. Solve for k: 6k=8, so k=86=43.

k=43; option (A).

Correct option: (C) x²2-x2-6.

Concept used. A quadratic with zeroes α,β can be written as k[x²-(α+β)x+αβ] for any non-zero constant k.

1. Find the sum and product of the zeroes: [] α+β=-3+4=1 [] αβ=(-3)(4)=-12
2. Form the basic polynomial: [] x²-(α+β)x+αβ=x²-x-12.
3. Any non-zero multiple works. Multiplying by k=12: [] 12(x²-x-12)=x²2-x2-6, which is option (C).

x²2-x2-6; option (C).

Correct option: (D) a=0, b=-6.

Concept used. For x²+(a+1)x+b, the sum of zeroes =-(a+1) and the product of zeroes =b.

1. Sum of zeroes: 2+(-3)=-1, and this equals -(a+1). [] -(a+1)=-1 ⇒ a+1=1 ⇒ a=0.
2. Product of zeroes: 2×(-3)=-6, and this equals b. [] b=-6.

a=0 and b=-6; option (D).

Correct option: (D) more than 3.

Concept used. Every quadratic with zeroes α,β has the form k[x²-(α+β)x+αβ], where k is any non-zero real number. Different values of k give different polynomials.

1. The base polynomial with zeroes -2 and 5 is x²-3x-10 (sum 3, product -10).
2. Multiplying by any non-zero k keeps the same zeroes: k(x²-3x-10).
3. Taking k=1,2,3,12,-7, gives endlessly many different polynomials, all with zeroes -2 and 5.
4. So the count is not 1, 2 or 3; it is unlimited.

There are infinitely many such polynomials, which is ``more than 3''; option (D).

Correct option: (B) ca.

Concept used. For a cubic ax³+bx²+cx+d with zeroes α,β,γ, the sum of products taken two at a time is αβ+γ+γα=ca.

1. Let the zeroes be α,β,γ with one of them zero, say γ=0.
2. Use the two-at-a-time relation: [] αβ+γ+γα=ca.
3. Put γ=0. The terms γ and γα vanish: [] αβ+0+0=ca.
4. So the product of the other two zeroes is αβ=ca.

Product of the other two zeroes =ca; option (B).

Correct option: (A) b-a+1.

Concept used. Since -1 is a zero, (x+1) is a factor, and the constant term of x³+ax²+bx+c equals -(product of all three zeroes). Combine this with the value of the polynomial at -1.

1. Let the zeroes be -1, β, γ. The required quantity is γ.
2. Product of all three zeroes (for x³+ax²+bx+c) is -c: [] (-1) γ=-c ⇒ γ=c.
3. Now use that -1 is a zero, so the polynomial is 0 at x=-1: [] (-1)³+a(-1)²+b(-1)+c=0 [] -1+a-b+c=0 ⇒ c=1-a+b.
4. Therefore γ=c=b-a+1.

Product of the other two zeroes =b-a+1; option (A).

Correct option: (B) both negative.

Concept used. The signs of the zeroes follow from the sum α+β=-ba and the product αβ=ca.

1. Here a=1, b=99, c=127.
2. Product of zeroes: αβ=ca=1271=127>0, so the two zeroes have the same sign.
3. Sum of zeroes: α+β=-ba=-99<0, so that common sign is negative.
4. (Both are real, since the discriminant 99²-4(127)=9801-508=9293>0.) Hence both zeroes are negative.

Both zeroes are negative; option (B).

Correct option: (A) cannot both be positive.

Concept used. For x²+kx+k, the sum of zeroes is -k and the product is k. Two positive zeroes would need a positive sum and a positive product at the same time.

1. Sum of zeroes =-k and product of zeroes =k.
2. For both zeroes to be positive, we would need sum >0 and product >0: that is -k>0 and k>0 together.
3. But -k>0 means k<0, while k>0 means k>0. These cannot hold at once for any k≠ 0.
4. So the zeroes can never both be positive. (When k<0 the product k<0, giving opposite signs; when k>0 the sum -k<0, so they are both negative if real.)

The zeroes can never both be positive; option (A).

Correct option: (C) c and a have the same sign.

Concept used. Two real zeroes are equal exactly when the discriminant b²-4ac=0.

1. Equal zeroes means b²-4ac=0.
2. Rearrange: 4ac=b².
3. The right side b²≥ 0, and since c≠ 0 we get b²>0, so 4ac>0, hence ac>0.
4. ac>0 means a and c have the same sign.

ac>0, so c and a have the same sign; option (C).

Correct option: (A) has no linear term and the constant term is negative.

Concept used. If the zeroes are α and -α, then the sum of zeroes is 0 (which kills the linear term) and the product is -α² (which is negative).

1. Let the zeroes be α and -α.
2. Sum of zeroes =α+(-α)=0. In x²+ax+b the sum is -a, so -a=0, i.e. a=0: there is no linear term.
3. Product of zeroes =α·(-α)=-α². In x²+ax+b the product is b, so b=-α².
4. Since α≠ 0 (otherwise both zeroes are 0 and not ``negatives'' in any useful sense), -α²<0, so the constant term b is negative.

The polynomial has no linear term and a negative constant term; option (A).

Correct option: (D).

Concept used. The graph of a quadratic polynomial is a parabola. A quadratic has at most 2 zeroes, so its graph can cross the x-axis in at most 2 points. A graph that meets the x-axis at 3 points cannot belong to a quadratic.

1. Recall: a degree-2 polynomial y=ax²+bx+c draws a single ∪ or ∩ shaped parabola.
2. A parabola can touch or cross the x-axis 0, 1, or 2 times, never more.
3. Graphs (A), (B) and (C) are parabolas (single bend), so each can be a quadratic.
4. Graph (D) is an S-shaped curve that crosses the x-axis at three points, which needs degree ≥ 3. So (D) is not the graph of a quadratic.

Graph (D) crosses the x-axis three times, so it is not a quadratic; option (D).

Verdict: No. x²-1 cannot be such a quotient.

Concept used. In the division algorithm p(x)=g(x) q(x)+r(x), the degree of the dividend equals deg g(x)+deg q(x) when the remainder has smaller degree. Degrees of the factors must add up correctly.

1. Here the dividend p(x)=x⁶+2x³+x-1 has degree 6, and the divisor g(x) has degree 5.
2. If the quotient were q(x)=x²-1 of degree 2, then deg(g(x) q(x))=5+2=7.
3. A degree-7 product cannot match a degree-6 dividend (the remainder has degree <5, so it cannot fix the mismatch).
4. For a degree-6 dividend and a degree-5 divisor, the quotient must have degree 6-5=1, not 2.

No, because the quotient of a degree-6 polynomial by a degree-5 polynomial must have degree 1, not 2.

Verdict / Answer: quotient =0 and remainder =ax²+bx+c.

Concept used. When the dividend has smaller degree than the divisor, division stops at once: the quotient is 0 and the whole dividend is the remainder.

1. The dividend ax²+bx+c has degree 2.
2. The divisor px³+qx²+rx+s has degree 3 (since p≠ 0).
3. Because deg(dividend)=2<3=deg(divisor), the divisor cannot ``go into'' the dividend even once.
4. So the quotient is 0, and the remainder is the dividend itself: [] ax²+bx+c=(px³+qx²+rx+s)· 0+(ax²+bx+c).

Quotient =0 and remainder =ax²+bx+c.

Verdict / Answer: deg p(x).

Concept used. A zero quotient means the divisor never divides into the dividend, which happens only when the dividend has the smaller degree.

1. The division algorithm gives p(x)=g(x) q(x)+r(x) with q(x)=0.
2. Then p(x)=g(x)· 0+r(x)=r(x), so p(x) is just the remainder.
3. The remainder must satisfy deg r(x).
4. Since p(x)=r(x), this means deg p(x).

The quotient is zero exactly when deg p(x).

Verdict / Answer: deg g(x)≤ deg p(x) (that is, deg p(x)≥ deg g(x)).

Concept used. A zero remainder means g(x) is a factor of p(x), so p(x)=g(x) q(x) and degrees add.

1. With remainder 0, the algorithm gives p(x)=g(x) q(x).
2. Taking degrees: deg p(x)=deg g(x)+deg q(x).
3. Since p(x) is non-zero, q(x) is non-zero, so deg q(x)≥ 0.
4. Therefore deg p(x)=deg g(x)+deg q(x)≥ deg g(x), i.e. g(x) has degree at most that of p(x).

deg g(x)≤ deg p(x), since g(x) is a factor of p(x).

Verdict: No. There is no odd integer k>1 giving equal zeroes.

Concept used. Equal zeroes need discriminant b²-4ac=0. Solve this for k and check whether any odd integer greater than 1 fits.

1. For x²+kx+k: a=1, b=k, c=k.
2. Equal zeroes require b²-4ac=0: [] k²-4(1)(k)=0 [] k²-4k=0 [] k(k-4)=0.
3. So k=0 or k=4.
4. Neither 0 nor 4 is an odd integer greater than 1. Hence no such odd k>1 exists.

No, because equal zeroes force k=0 or k=4, and neither is an odd integer greater than 1.

Verdict: False.

Concept used. For two positive zeroes, the sum -ba>0 and the product ca>0. The sign of b then comes out opposite to that of a.

1. Both zeroes positive ⇒ sum >0: -ba>0, so ba<0, meaning a and b have opposite signs.
2. Both zeroes positive ⇒ product >0: ca>0, so a and c have the same sign.
3. So a and c match in sign, but b is opposite. They cannot all three share one sign.
4. Example: x²-5x+6=(x-2)(x-3) has positive zeroes 2,3, with a=1>0, b=-5<0, c=6>0 – not all the same sign.

False; the middle coefficient b always has the opposite sign to a and c.

Verdict: False.

Concept used. A quadratic with equal zeroes touches the x-axis at exactly one point. So a single x-axis meeting point is fully consistent with a quadratic.

1. A quadratic can have two distinct zeroes, two equal zeroes, or no real zeroes.
2. When the two zeroes are equal, the parabola just touches the x-axis at one point (the vertex sits on the axis).
3. Example: y=x²-4x+4=(x-2)² meets the x-axis only at x=2, yet it is a quadratic.
4. So ``one intersection point'' does not rule out a quadratic.

False; a quadratic with equal zeroes, such as (x-2)², meets the x-axis at exactly one point.

Verdict: True.

Concept used. The number of x-axis crossings equals the number of real zeroes, but a high-degree polynomial can also have exactly two real zeroes. So two crossings do not force degree 2.

1. ``Exactly two intersection points'' means the polynomial has exactly two real zeroes.
2. A quadratic has at most two zeroes, so it can fit, but it is not the only possibility.
3. Higher-degree polynomials can have exactly two real zeroes too. For example y=x⁴-1 cuts the x-axis only at x=1 and x=-1 (its other two zeroes are not real), yet it has degree 4.
4. So a two-crossing graph need not be a quadratic.

True; for example x⁴-1 meets the x-axis at exactly two points but has degree 4.

Verdict: True.

Concept used. A cubic with zeroes α,β,γ is a(x-α)(x-β)(x-γ). Setting two zeroes to 0 shows which terms survive.

1. Let the zeroes be 0, 0, γ. The cubic is [] a(x-0)(x-0)(x-γ)=a x²(x-γ).
2. Expand: a x²(x-γ)=a x³-aγ x².
3. The result has only an x³ term and an x² term.
4. There is no x (linear) term and no constant term.

True; with two zeroes equal to 0 the cubic is ax³-aγ x², which has no linear or constant term.

Verdict: True.

Concept used. Write the cubic as a(x-α)(x-β)(x-γ) with all zeroes negative. Then each factor (x+|α|) has positive coefficients, so the product has all coefficients of one sign.

1. Let the negative zeroes be -p,-q,-r with p,q,r>0. The cubic is [] a(x+p)(x+q)(x+r).
2. Multiply out the three factors: [] (x+p)(x+q)(x+r)=x³+(p+q+r)x²+(pq+qr+rp)x+pqr.
3. Each of p+q+r, pq+qr+rp and pqr is positive, so all four coefficients inside are positive.
4. Multiplying by a keeps them all of the same sign as a. Hence every coefficient and the constant term share one sign.

True; all the coefficients and the constant term come out with the same sign as the leading coefficient a.

Verdict: False.

Concept used. Match x³+ax²-bx+c with the zero–coefficient relations. With all zeroes positive, each of a, b, c turns out strictly negative.

1. Let the positive zeroes be α,β,γ>0. Compare with x³+ax²-bx+c: [] sum =α+β+γ=-a, so a=-(α+β+γ)<0. [] pair-sum =αβ+γ+α=-b, so b=-(αβ+γ+α)<0. [] product =αγ=-c, so c=-αγ<0.
2. All of a, b, c are strictly negative, so none of them is non-negative.
3. Therefore the claim ``at least one is non-negative'' is false.

False; with all zeroes positive, a, b and c are all negative.

Verdict: False.

Concept used. Equal zeroes need discriminant b²-4ac=0. Solving this for k gives two values, not one.

1. For kx²+x+k: a=k, b=1, c=k (and k≠ 0).
2. Equal zeroes require b²-4ac=0: [] 1²-4(k)(k)=0 [] 1-4k²=0 [] k²=14.
3. So k=12 or k=-12.
4. There are two values, so 12 is not the only one.

False; equal zeroes occur for k=12 and k=-12.

Concept used. Split the middle term so the two parts multiply to a× c and add to b; then read the zeroes from the factors and check them against α+β=-ba and αβ=ca.

1. Here a=4, b=-3, c=-1, so a× c=4×(-1)=-4. Find two numbers that multiply to -4 and add to -3: these are -4 and +1.
2. Split the middle term: [] 4x²-3x-1=4x²-4x+x-1.
3. Group and factor: [] =4x(x-1)+1(x-1)=(x-1)(4x+1).
4. Set each factor to 0: x-1=0⇒ x=1, and 4x+1=0⇒ x=-14.
5. Verify the sum: [] α+β=1+(-14)=34 [] -ba=--34=34   
6. Verify the product: [] αβ=1×(-14)=-14 [] ca=-14=-14   

Zeroes are 1 and -14; both relations check out.

Concept used. Split the middle term using a× c, factor, then check α+β=-ba and αβ=ca.

1. Here a=3, b=4, c=-4, so a× c=3×(-4)=-12. Two numbers with product -12 and sum 4 are 6 and -2.
2. Split the middle term: [] 3x²+4x-4=3x²+6x-2x-4.
3. Group and factor: [] =3x(x+2)-2(x+2)=(x+2)(3x-2).
4. Zeroes: x+2=0⇒ x=-2, and 3x-2=0⇒ x=23.
5. Verify the sum: [] α+β=-2+23=-43 [] -ba=-43   
6. Verify the product: [] αβ=(-2)×23=-43 [] ca=-43=-43   

Zeroes are -2 and 23; both relations check out.

Concept used. Split the middle term using a× c, factor, then verify α+β=-ba and αβ=ca.

1. Here a=5, b=12, c=7, so a× c=5× 7=35. Two numbers with product 35 and sum 12 are 7 and 5.
2. Split the middle term: [] 5t²+12t+7=5t²+7t+5t+7.
3. Group and factor: [] =t(5t+7)+1(5t+7)=(5t+7)(t+1).
4. Zeroes: 5t+7=0⇒ t=-75, and t+1=0⇒ t=-1.
5. Verify the sum: [] α+β=-75+(-1)=-125 [] -ba=-125   
6. Verify the product: [] αβ=(-75)×(-1)=75 [] ca=75   

Zeroes are -75 and -1; both relations check out.

Concept used. First take out the common factor t, then split the remaining quadratic. For a cubic at³+bt²+ct+d, check α=-ba, αβ=ca and αγ=-da.

1. Take out the common t: [] t³-2t²-15t=t(t²-2t-15).
2. Factor the quadratic t²-2t-15: product -15, sum -2 gives -5 and +3: [] t²-2t-15=(t-5)(t+3).
3. So t³-2t²-15t=t(t-5)(t+3), and the zeroes are 0, 5, -3.
4. Here a=1, b=-2, c=-15, d=0.
5. Verify the sum: [] α+β+γ=0+5+(-3)=2 [] -ba=--21=2   
6. Verify the pair-sum: [] αβ+γ+α=(0)(5)+(5)(-3)+(-3)(0)=-15 [] ca=-151=-15   
7. Verify the product: [] αγ=0× 5×(-3)=0 [] -da=-01=0   

Zeroes are 0, 5 and -3; all three cubic relations check out.

Concept used. Clear the fractions first by multiplying through, factor the cleaner polynomial (the zeroes are unchanged), then verify the relations using the original a,b,c.

1. Multiply the polynomial by 4 to clear fractions (this scales the polynomial but keeps the same zeroes): [] 4(2x²+72x+34)=8x²+14x+3.
2. Factor 8x²+14x+3: product 8× 3=24, sum 14 gives 12 and 2: [] 8x²+14x+3=8x²+12x+2x+3 [] =4x(2x+3)+1(2x+3)=(2x+3)(4x+1).
3. Zeroes: 2x+3=0⇒ x=-32, and 4x+1=0⇒ x=-14.
4. Now use the original coefficients a=2, b=72, c=34.
5. Verify the sum: [] α+β=-32+(-14) =-74 [] -ba=-7/22=-74   
6. Verify the product: [] αβ=(-32)× (-14)=38 [] ca=3/42=38   

Zeroes are -32 and -14; both relations check out.

Concept used. Even with surd coefficients, split the middle term so the two parts multiply to a× c and add to b; then verify α+β=-ba and αβ=ca.

1. Here a=4, b=5√2, c=-3, so a× c=4×(-3)=-12. We need two terms whose product is -12x² and whose sum is 5√2 x: these are 6√2 x and -√2 x (since 6√2·(-√2)=-12 and 6√2-√2=5√2).
2. Split the middle term: [] 4x²+5√2 x-3=4x²+6√2 x-√2 x-3.
3. Group and factor: [] =2√2 x(√2 x+3)-1(√2 x+3) =(√2 x+3)(2√2 x-1).
4. Zeroes: √2 x+3=0⇒ x=-3√2=-3√22, and 2√2 x-1=0⇒ x=12√2=√24.
5. Verify the sum: [] α+β=-3√22+√24 =-6√24+√24=-5√24 [] -ba=-5√24   
6. Verify the product: [] αβ=(-3√22) ×√24=-3× 28=-34 [] ca=-34=-34   

Zeroes are -3√22 and √24; both relations check out.

Concept used. Split the middle coefficient -(1+2√2) into two parts whose product is a× c and whose sum is the middle coefficient; then verify the sum and product of zeroes.

1. Here a=2, b=-(1+2√2), c=√2, so a× c=2√2.
2. Split -(1+2√2)s=-s-2√2 s, because (-1)×(-2√2)=2√2 matches a× c and -1-2√2 matches b.
3. Write it out: [] 2s²-(1+2√2)s+√2=2s²-2√2 s-s+√2.
4. Group and factor: [] =2s(s-√2)-1(s-√2)=(s-√2)(2s-1).
5. Zeroes: s-√2=0⇒ s=√2, and 2s-1=0⇒ s=12.
6. Verify the sum: [] α+β=√2+12 [] -ba=--(1+2√2)2 =1+2√22=12+√2   
7. Verify the product: [] αβ=√2×12=√22 [] ca=√22   

Zeroes are √2 and 12; both relations check out.

Concept used. Split the middle term using a× c with a surd present; factor by grouping, then verify α+β=-ba and αβ=ca.

1. Here a=1, b=4√3, c=-15, so a× c=-15. We need two surd terms with product -15v² and sum 4√3 v: these are 5√3 v and -√3 v (since 5√3·(-√3)=-5× 3=-15 and 5√3-√3=4√3).
2. Split the middle term: [] v²+4√3 v-15=v²+5√3 v-√3 v-15.
3. Group and factor: [] =v(v+5√3)-√3(v+5√3)=(v+5√3)(v-√3).
4. Zeroes: v+5√3=0⇒ v=-5√3, and v-√3=0⇒ v=√3.
5. Verify the sum: [] α+β=-5√3+√3=-4√3 [] -ba=-4√31=-4√3   
6. Verify the product: [] αβ=(-5√3)(√3)=-5× 3=-15 [] ca=-151=-15   

Zeroes are -5√3 and √3; both relations check out.

Concept used. Clear the fraction by multiplying through (the zeroes are unchanged), split the middle term, factor, then verify with the original coefficients.

1. Multiply by 2 to clear the fraction: [] 2(y²+32√5 y-5) =2y²+3√5 y-10.
2. For 2y²+3√5 y-10: a× c=2×(-10)=-20. We need two surd terms with product -20y² and sum 3√5 y: these are 4√5 y and -√5 y (since 4√5·(-√5)=-4× 5=-20 and 4√5-√5=3√5).
3. Split and group: [] 2y²+4√5 y-√5 y-10 [] =2y(y+2√5)-√5(y+2√5) =(y+2√5)(2y-√5).
4. Zeroes: y+2√5=0⇒ y=-2√5, and 2y-√5=0⇒ y=√52.
5. Use the original coefficients a=1, b=32√5, c=-5.
6. Verify the sum: [] α+β=-2√5+√52 =-4√52+√52=-3√52 [] -ba=-3√5/21=-3√52   
7. Verify the product: [] αβ=(-2√5)×√52 =-2× 52=-5 [] ca=-51=-5   

Zeroes are -2√5 and √52; both relations check out.

Concept used. Clear the fractions by multiplying through, split the middle term of the integer polynomial, factor, then verify with the original coefficients.

1. Multiply by 3 to clear fractions: [] 3(7y²-113y-23) =21y²-11y-2.
2. For 21y²-11y-2: a× c=21×(-2)=-42. Two numbers with product -42 and sum -11 are -14 and +3.
3. Split and group: [] 21y²-14y+3y-2 [] =7y(3y-2)+1(3y-2)=(3y-2)(7y+1).
4. Zeroes: 3y-2=0⇒ y=23, and 7y+1=0⇒ y=-17.
5. Use the original coefficients a=7, b=-113, c=-23.
6. Verify the sum: [] α+β=23+(-17) =1421-321=1121 [] -ba=--11/37=1121   
7. Verify the product: [] αβ=23×(-17) =-221 [] ca=-2/37=-221   

Zeroes are 23 and -17; both relations check out.

Concept used. A quadratic with zeroes summing to S and multiplying to P is k[x²-Sx+P]. Choosing k to clear denominators gives a tidy polynomial; then factorise it.

1. Here S=-83 and P=43. The basic polynomial is [] x²-Sx+P=x²+83x+43.
2. Multiply by k=3 to clear fractions (this keeps the same zeroes): [] 3x²+8x+4.
3. Factorise 3x²+8x+4: product 3× 4=12, sum 8 gives 6 and 2: [] 3x²+8x+4=3x²+6x+2x+4 [] =3x(x+2)+2(x+2)=(x+2)(3x+2).
4. Zeroes: x+2=0⇒ x=-2, and 3x+2=0⇒ x=-23.
5. Check: sum =-2-23=-83=S and product =(-2)(-23)=43=P.

One such polynomial is 3x²+8x+4, with zeroes -2 and -23.

Concept used. Use x²-Sx+P, scale to clear denominators, then factorise by splitting the middle term.

1. Here S=218 and P=516. The basic polynomial is [] x²-218x+516.
2. Multiply by k=16 to clear fractions: [] 16x²-42x+5.
3. Factorise 16x²-42x+5: product 16× 5=80, sum -42 gives -40 and -2: [] 16x²-42x+5=16x²-40x-2x+5 [] =8x(2x-5)-1(2x-5)=(2x-5)(8x-1).
4. Zeroes: 2x-5=0⇒ x=52, and 8x-1=0⇒ x=18.
5. Check: sum =52+18=20+18 =218=S and product =52×18=516=P.

One such polynomial is 16x²-42x+5, with zeroes 52 and 18.

Concept used. Use x²-Sx+P directly (no fractions to clear here), then split the middle term using a surd.

1. Here S=-2√3 and P=-9. The polynomial is [] x²-(-2√3)x+(-9)=x²+2√3 x-9.
2. Factorise x²+2√3 x-9: a× c=-9. We need two surd terms with product -9x² and sum 2√3 x: these are 3√3 x and -√3 x (since 3√3·(-√3)=-3× 3=-9 and 3√3-√3=2√3).
3. Split and group: [] x²+3√3 x-√3 x-9 [] =x(x+3√3)-√3(x+3√3) =(x+3√3)(x-√3).
4. Zeroes: x+3√3=0⇒ x=-3√3, and x-√3=0⇒ x=√3.
5. Check: sum =-3√3+√3=-2√3=S and product =(-3√3)(√3)=-9=P.

One such polynomial is x²+2√3 x-9, with zeroes -3√3 and √3.

Concept used. Use x²-Sx+P. Rationalise and scale to get whole or surd-integer coefficients, then split the middle term.

1. Here S=-32√5 and P=-12. Rationalise S: [] S=-32√5=-3√52× 5 =-3√510.
2. Basic polynomial: [] x²-Sx+P=x²+3√510x-12.
3. Multiply by k=10 to clear fractions: [] 10x²+3√5 x-5.
4. Factorise: a× c=10×(-5)=-50. We need two surd terms with product -50x² and sum 3√5 x: these are 5√5 x and -2√5 x (since 5√5·(-2√5)=-10× 5=-50 and 5√5-2√5=3√5).
5. Split and group: [] 10x²+5√5 x-2√5 x-5 [] =5x(2x+√5)-√5(2x+√5) =(2x+√5)(5x-√5).
6. Zeroes: 2x+√5=0⇒ x=-√52, and 5x-√5=0⇒ x=√55=1√5.
7. Check: product =(-√52) ×√55=-510=-12=P.

One such polynomial is 10x²+3√5 x-5, with zeroes -√52 and √55.

Concept used. For x³-6x²+3x+10 the sum of zeroes is -ba=6 and the product of zeroes is -da=-10. Use the arithmetic-progression form to turn these into equations.

1. The three zeroes are a, a+b, a+2b (an AP with common difference b).
2. Sum of zeroes =6: [] a+(a+b)+(a+2b)=6 [] 3a+3b=6 [] a+b=2.   (So the middle zero a+b=2.)
3. Product of zeroes =-10: [] a(a+b)(a+2b)=-10.
4. Substitute a+b=2, so a=2-b and a+2b=2+b: [] (2-b)(2)(2+b)=-10 [] 2(4-b²)=-10 [] 4-b²=-5 [] b²=9⇒ b=± 3.
5. If b=3: a=2-3=-1. If b=-3: a=2-(-3)=5.
6. Both give the same set of zeroes: for a=-1,b=3 they are -1, 2, 5; for a=5,b=-3 they are 5, 2, -1.

a=-1, b=3 (or a=5, b=-3); the zeroes are -1, 2 and 5.

Concept used. If √2 is a zero, then (x-√2) is a factor. Divide the cubic by (x-√2) to get a quadratic, then factorise that quadratic for the other two zeroes.

1. Divide 6x³+√2 x²-10x-4√2 by (x-√2).
2. First term: 6x³x=6x²; 6x²(x-√2)=6x³-6√2 x². Subtract: [] remainder so far =(√2+6√2)x²-10x-4√2 =7√2 x²-10x-4√2.
3. Next term: 7√2 x²x=7√2 x; 7√2 x(x-√2)=7√2 x²-14x. Subtract: [] remainder so far =(-10+14)x-4√2=4x-4√2.
4. Last term: 4xx=4; 4(x-√2)=4x-4√2. Subtract: remainder =0.
5. So the quotient is 6x²+7√2 x+4, and [] 6x³+√2 x²-10x-4√2 =(x-√2)(6x²+7√2 x+4).
6. Factorise 6x²+7√2 x+4: a× c=24. Two surd terms with product 24x² and sum 7√2 x are 3√2 x and 4√2 x (since 3√2· 4√2=12× 2=24 and 3√2+4√2=7√2): [] 6x²+3√2 x+4√2 x+4 [] =3x(2x+√2)+2√2(2x+√2) =(2x+√2)(3x+2√2).
7. Other zeroes: 2x+√2=0⇒ x=-√22, and 3x+2√2=0⇒ x=-2√23.

The other two zeroes are -√22 and -2√23.

Concept used. If x²+2x+k is a factor, then on dividing 2x⁴+x³-14x²+5x+6 by it the remainder must be 0. Set the remainder to 0 to find k.

1. Divide 2x⁴+x³-14x²+5x+6 by x²+2x+k. Carrying out polynomial long division gives quotient 2x²-3x-(8+2k) and [] remainder =(7k+21)x+(6+k(8+2k)) =(7k+21)x+(2k²+8k+6).
2. For an exact factor the remainder is the zero polynomial, so both coefficients vanish.
3. Coefficient of x: 7k+21=0⇒ k=-3.
4. Constant term (check with k=-3): 2(-3)²+8(-3)+6=18-24+6=0. Both conditions agree on k=-3.
5. With k=-3, the factor is x²+2x-3=(x+3)(x-1), whose zeroes are -3 and 1.
6. The quotient becomes 2x²-3x-(8+2(-3))=2x²-3x-2. Factorise: 2x²-3x-2=2x²-4x+x-2=2x(x-2)+1(x-2)=(x-2)(2x+1), zeroes 2 and -12.
7. So 2x⁴+x³-14x²+5x+6=(x²+2x-3)(2x²-3x-2), and its four zeroes are -3, 1, 2, -12.

k=-3. Zeroes of x²+2x-3 are -3, 1; zeroes of 2x⁴+x³-14x²+5x+6 are -3, 1, 2, -12.

Concept used. Since x-√5 is a factor, divide the cubic by it to get a quadratic, then solve the quadratic (here by the quadratic formula) for the remaining zeroes.

1. Divide x³-3√5 x²+13x-3√5 by (x-√5).
2. First term x²: x²(x-√5)=x³-√5 x². Subtract: [] (-3√5+√5)x²+13x-3√5 =-2√5 x²+13x-3√5.
3. Next term -2√5 x: -2√5 x(x-√5)=-2√5 x²+10x. Subtract: [] (13-10)x-3√5=3x-3√5.
4. Last term 3: 3(x-√5)=3x-3√5. Subtract: remainder =0.
5. Quotient =x²-2√5 x+3, so [] x³-3√5 x²+13x-3√5 =(x-√5)(x²-2√5 x+3).
6. Solve x²-2√5 x+3=0 by the quadratic formula x=-B±√B²-4AC2A with A=1, B=-2√5, C=3: [] x=2√5±√(2√5)²-4(1)(3)2 [] =2√5±√20-122 =2√5±√82 [] =2√5± 2√22=√5±√2.
7. So the other two zeroes are √5+√2 and √5-√2.

All three zeroes are √5, √5+√2 and √5-√2.

Concept used. If every zero of q(x) is also a zero of p(x), then q(x) is a factor of p(x). Dividing p(x) by q(x) and forcing the remainder to be 0 fixes a and b.

1. Divide p(x)=x⁵-x⁴-4x³+3x²+3x+b by q(x)=x³+2x²+0· x+a.
2. First term x²: x² q(x)=x⁵+2x⁴+ax². Subtract: [] (-1-2)x⁴-4x³+(3-a)x²+3x+b =-3x⁴-4x³+(3-a)x²+3x+b.
3. Next term -3x: -3x q(x)=-3x⁴-6x³-3ax. Subtract: [] (-4+6)x³+(3-a)x²+(3+3a)x+b =2x³+(3-a)x²+(3+3a)x+b.
4. Next term 2: 2 q(x)=2x³+4x²+2a. Subtract: [] remainder =(3-a-4)x²+(3+3a)x+(b-2a) =(-a-1)x²+(3+3a)x+(b-2a).
5. For q(x) to be a factor, every remainder coefficient is 0: [] -a-1=0⇒ a=-1; [] 3+3a=0⇒ a=-1 (consistent); [] b-2a=0⇒ b=2a=2(-1)=-2.
6. So a=-1 and b=-2. Then q(x)=x³+2x²-1 and the quotient is x²-3x+2=(x-1)(x-2).
7. The quotient's zeroes 1 and 2 are the zeroes of p(x) that come from the extra factor, not from q(x). (Check: q(1)=1+2-1=2≠ 0 and q(2)=8+8-1=15≠ 0, so 1 and 2 are genuinely not zeroes of q(x).)

a=-1 and b=-2. The zeroes 1 and 2 of p(x) are not zeroes of q(x).