Answer: \(a=2\), \(b=0\).

Since \(x-3\) is a factor of \(\mathrm{p}(x)\), the factor theorem gives

\(\mathrm{p}(3)=0\).

Substitute \(x=3\) into

\(\mathrm{p}(x)=ax^3-7x^2-bx+9\):

\(27a-7(9)-3b+9=0\).

This simplifies to

\(27a-3b-54=0\).

Dividing by \(3\),

\(9a-b=18\). \(\quad\text{(1)}\)

When \(\mathrm{p}(x)\) is divided by \(x+2\), the remainder is \(\mathrm{p}(-2)\). The remainder is given as \(-35\), so

\(\mathrm{p}(-2)=-35\).

Substituting \(x=-2\),

\(-8a-7(4)+2b+9=-35\).

So

\(-8a+2b-19=-35\),

which gives

\(-8a+2b=-16\).

Dividing by \(2\),

\(-4a+b=-8\). \(\quad\text{(2)}\)

Add equations (1) and (2):

\((9a-b)+(-4a+b)=18-8\).

Thus

\(5a=10\),

so

\(a=2\).

Substitute into \(-4a+b=-8\):

\(-4(2)+b=-8\),

so

\(b=0\).

Answer: \(a=2\), \(b=-1\).

Since \(x+2\) is a factor of \(\mathrm p(x)\), the factor theorem gives

\(\mathrm p(-2)=0.\)

Substitute \(x=-2\) into \(\mathrm p(x)=x^3+ax^2+bx-2\):

\((-2)^3+a(-2)^2+b(-2)-2=0.\)

So

\(-8+4a-2b-2=0,\)

which simplifies to

\(4a-2b=10.\)

Divide by \(2\):

\(2a-b=5. \tag{1}\)

By the remainder theorem, the remainder when \(\mathrm p(x)\) is divided by \(x-3\) is \(\mathrm p(3)\). This remainder is \(40\), so

\(3^3+a(3)^2+b(3)-2=40.\)

Therefore

\(27+9a+3b-2=40,\)

so

\(9a+3b=15.\)

Divide by \(3\):

\(3a+b=5. \tag{2}\)

From (1),

\(b=2a-5.\)

Substitute this into (2):

\(3a+(2a-5)=5.\)

So

\(5a=10,\)

giving

\(a=2.\)

Then

\(b=2(2)-5=-1.\)

Therefore

\(a=2,\qquad b=-1.\)

Answer: \(a=7\), \(b=14\); \(\mathrm{p}(x)=0\) has only one real root.

(a) Since \(x+2\) is a factor of \(\mathrm p(x)\), the factor theorem gives

\(\mathrm p(-2)=0.\)

Substitute \(x=-2\) into \(\mathrm p(x)=2x^3+ax^2+13x+b\):

\(2(-2)^3+a(-2)^2+13(-2)+b=0.\)

So

\(-16+4a-26+b=0,\)

which gives

\(4a+b=42. \tag{1}\)

The remainder when \(\mathrm p(x)\) is divided by \(x+1\) is \(\mathrm p(-1)\). This remainder is \(6\), so

\(2(-1)^3+a(-1)^2+13(-1)+b=6.\)

Hence

\(-2+a-13+b=6,\)

so

\(a+b=21. \tag{2}\)

Subtracting (2) from (1),

\(3a=21,\)

so \(a=7\). Then from \(a+b=21\),

\(b=14.\)

(b) Therefore

\(\mathrm p(x)=2x^3+7x^2+13x+14.\)

Since \(x+2\) is a factor, divide by \(x+2\):

\(\mathrm p(x)=(x+2)(2x^2+3x+7).\)

Thus \(x=-2\) is one real root. The other possible roots come from

\(2x^2+3x+7=0.\)

The discriminant is

\(3^2-4(2)(7)=9-56=-47.\)

Since this is negative, the quadratic has no real roots. Hence \(\mathrm p(x)=0\) has only one real root.

Answer: \(a=2\), \(b=12\), \(c=-9\).

Use the factor theorem and the remainder theorem to turn the given information into equations for the unknown constants. After finding the constants, factorise the polynomial if required.

\(\mathrm p(x)=ax^3+11x^2+bx+c\), so

\(\mathrm p'(x)=3ax^2+22x+b\).

Since \(\mathrm p'(0)=12\), we get \(b=12\).

Since \(x+3\) is a factor, \(\mathrm p(-3)=0\):

\(-27a+99-3b+c=0\).

Using \(b=12\), this becomes

\(-27a+63+c=0\).

The remainder when \(\mathrm p(x)\) is divided by \(x-1\) is \(\mathrm p(1)\), so

\(a+11+b+c=16\).

Using \(b=12\),

\(a+c=-7\).

From \(-27a+63+c=0\), \(c=27a-63\). Substitute into \(a+c=-7\):

\(a+27a-63=-7\), so \(28a=56\), and \(a=2\).

Then \(c=-9\).

This gives the required answer: \(a=2\), \(b=12\), \(c=-9\).

Answer: \(a=6\), \(b=1\), \(c=6\).

Use the factor theorem and the remainder theorem to turn the given information into equations for the unknown constants. After finding the constants, factorise the polynomial if required.

Since \(x+2\) is a factor, \(\mathrm p(-2)=0\):

\(-8a+4b+38+c=0.\)

When \(\mathrm p(x)\) is divided by \(x+1\), the remainder is \(\mathrm p(-1)\), so

\(-a+b+19+c=20.\)

Subtracting these equations eliminates \(c\):

\(-8a+4b+38+c)-(-a+b+19+c)=0-20\).

This gives \(-7a+3b+19=-20\), so \(7a-3b=39\), as required.

Now \(\mathrm p'(x)=3ax^2+2bx-19\). Since the remainder when \(\mathrm p'(x)\) is divided by \(x-1\) is \(1\),

\(\mathrm p'(1)=1\).

Thus \(3a+2b-19=1\), so \(3a+2b=20\).

Solve

\(7a-3b=39,\quad 3a+2b=20.\)

This gives \(a=6\), \(b=1\).

Using \(-a+b+19+c=20\),

\(-6+1+19+c=20\), so \(c=6\).

This gives the required answer: \(a=6\), \(b=1\), \(c=6\).

Answer: \(a=6\), \(b=-14\), \(c=-8\), and \(p(x)=(3x-4)(2x+1)(x+2)\).

Use the remainder theorem: when a polynomial \(P(x)\) is divided by \(x-a\), the remainder is \(P(a)\). If \(x-a\) is a factor, then \(P(a)=0\).

(a) Differentiate twice:

Given \(p''\left(\frac12\right)=32\),

\(6a\left(\frac12\right)+14=32.\)

So

\(3a=18,\)

and therefore

\(a=6.\)

(b) Since \(3x-4\) is a factor, \(x=\frac43\) is a root. With \(a=6\),

\(p(x)=6x^3+7x^2+bx+c.\)

So

\(p\left(\frac43\right)=0.\)

Multiplying by \(9\) after substitution gives

\(128+112+12b+9c=0,\)

so

\(12b+9c=-240.\)

Dividing by \(3\),

\(4b+3c=-80. \tag{1}\)

The remainder when \(p(x)\) is divided by \(x+1\) is \(p(-1)\). Hence

\(p(-1)=7.\)

Thus

\(-6+7-b+c=7,\)

so

\(-b+c=6. \tag{2}\)

From (2), \(c=b+6\). Substitute this into (1):

\(4b+3(b+6)=-80.\)

So

\(7b=-98,\)

giving

(c) Therefore

\(p(x)=6x^3+7x^2-14x-8.\)

Dividing by \(3x-4\) gives

\(p(x)=(3x-4)(2x^2+5x+2).\)

(d) Factorise the quadratic:

\(2x^2+5x+2=(2x+1)(x+2).\)

Hence

\(p(x)=(3x-4)(2x+1)(x+2).\)

Answer: \(a=6\), \(b=-10\), \(c=0\); \(P(x)=x(3x+2)(2x-5)\).

Use the remainder theorem: when a polynomial \(P(x)\) is divided by \(x-a\), the remainder is \(P(a)\). If \(x-a\) is a factor, then \(P(a)=0\).

Since \(P(x)\) is divisible by \(x\), we have \(P(0)=0\). Therefore

\(c=0\).

The remainder when \(P(x)\) is divided by \(2x+1\) is \(P\left(-\frac12\right)\), so

\(P\left(-\frac12\right)=\frac32\).

Using \(c=0\),

\(-\frac{a}{8}-\frac{11}{4}-\frac{b}{2}=\frac32\).

Multiply by \(-8\):

\(a+4b=-34\).

Now differentiate:

\(P'(x)=3ax^2-22x+b\).

Given \(P'(2)=18\),

\(12a-44+b=18\),

so

\(12a+b=62\).

Solve the simultaneous equations

\(a+4b=-34\)

and

\(12a+b=62\).

From \(a=-34-4b\). Substitute:

\(12(-34-4b)+b=62\).

So \(-408-47b=62\), giving \(b=-10\).

Then \(a+4(-10)=-34\), so \(a=6\).

Thus \(a=6\), \(b=-10\), and \(c=0\).

Now

\(P(x)=6x^3-11x^2-10x=x(6x^2-11x-10)\).

Factorise the quadratic:

\(6x^2-11x-10=(3x+2)(2x-5)\).

Therefore

\(P(x)=x(3x+2)(2x-5)\).

Answer: (a) \(a=6\), \(b=7\), \(c=12\); (b) \(p(x)=(3x-1)(2x^2+3x+5)\), so the only real root is \(x=\frac13\).

Use the remainder theorem: when a polynomial \(P(x)\) is divided by \(x-a\), the remainder is \(P(a)\). If \(x-a\) is a factor, then \(P(a)=0\).

(a) Since

\(p(x)=ax^3+bx^2+cx-5\),

we have

\(p'(x)=3ax^2+2bx+c\).

Given \(p'(0)=12\), it follows that

\(c=12\).

Since \(3x-1\) is a factor, \(x=\frac13\) is a root. Therefore

\(p\left(\frac13\right)=0\).

Using \(c=12\),

\(\frac{a}{27}+\frac{b}{9}+\frac{12}{3}-5=0\).

So

\(\frac{a}{27}+\frac{b}{9}=1\),

which gives

\(a+3b=27\).

The remainder when divided by \(x-2\) is \(p(2)\). Since the remainder is 95,

\(p(2)=95\).

So

\(8a+4b+2(12)-5=95\).

Hence

\(8a+4b+19=95\),

so

\(2a+b=19\).

Solve

\(a+3b=27\)

and

\(2a+b=19\).

From the second equation, \(b=19-2a\). Substitute into the first:

\(a+3(19-2a)=27\).

So \(a+57-6a=27\), giving \(-5a=-30\), and hence \(a=6\).

Then \(2(6)+b=19\), so \(b=7\).

Therefore \(a=6\), \(b=7\), and \(c=12\).

(b) Now

\(p(x)=6x^3+7x^2+12x-5\).

Since \(3x-1\) is a factor, divide or factorise to get

\(p(x)=(3x-1)(2x^2+3x+5)\).

The quadratic factor has discriminant

\(3^2-4(2)(5)=9-40=-31\).

Since the discriminant is negative, \(2x^2+3x+5=0\) has no real roots.

Therefore the only real root of \(p(x)=0\) is from \(3x-1=0\), which gives \(x=\frac13\).

Hence \(p(x)=0\) has only one real root.

Answer: (a) \(a=6,\ b=1\); (b) \(\mathrm P(x)=(2x+1)(3x^2-x+2)\), and the quadratic factor has negative discriminant.

Use the remainder theorem: when a polynomial \(P(x)\) is divided by \(x-a\), the remainder is \(P(a)\). If \(x-a\) is a factor, then \(P(a)=0\).

(a) Since \(2x+1\) is a factor, \(x=-\frac12\) is a root.

So

\(\mathrm P\left(-\frac12\right)=0.\)

This gives

\(-\frac{a}{8}+\frac{b}{4}-\frac32+2=0.\)

Hence

\(-a+2b+4=0.\)

The remainder when divided by \(x+1\) is \(\mathrm P(-1)\), so

\(\mathrm P(-1)=-6.\)

Therefore

\(-a+b-3+2=-6,\)

which gives

\(-a+b+5=0.\)

Solving

\(-a+2b+4=0,\quad -a+b+5=0\)

gives

\(b=1,\quad a=6.\)

(b) With these values,

\(\mathrm P(x)=6x^3+x^2+3x+2.\)

Since \(2x+1\) is a factor, divide by \(2x+1\):

\(\mathrm P(x)=(2x+1)(3x^2-x+2).\)

The quadratic factor has discriminant

\((-1)^2-4(3)(2)=1-24=-23.\)

Since this is negative, \(3x^2-x+2\) has no real roots.

The only real root therefore comes from \(2x+1=0\), so \(\mathrm P(x)=0\) has only one real root.

Answer: The remainder is \(-24\).

Answer: The remainder is \(-24\).

Since \(x-3\) is a factor, the factor theorem gives

\(p(3)=0\).

So

\(27m-17(9)+3n+6=0\),

which simplifies to

\(27m+3n=147\),

or

\(9m+n=49\). \(\quad (1)\)

When \(p(x)\) is divided by \(x+1\), the remainder is \(p(-1)\). Hence

\(p(-1)=-12\).

Therefore

\(-m-17-n+6=-12\),

so

\(m+n=1\). \(\quad (2)\)

Subtracting (2) from (1) gives

\(8m=48\),

so \(m=6\). Then \(n=1-6=-5\).

The remainder when \(p(x)\) is divided by \(x-2\) is \(p(2)\):

\(p(2)=6(8)-17(4)-5(2)+6\).

Thus

\(p(2)=48-68-10+6=-24\).

Answer: \(a=6,\ b=-3\), \(\mathrm p(x)=(x-2)(6x^2+3x+3)\), one real solution only.

Since \(x-2\) is a factor,

\(\mathrm p(2)=0.\)

So

\(8a-36+2b-6=0,\)

which gives

\(8a+2b=42. \tag{1}\)

The remainder when divided by \(x-3\) is \(66\), so

\(\mathrm p(3)=66.\)

Hence

\(27a-81+3b-6=66,\)

so

\(27a+3b=153. \tag{2}\)

Divide (1) by \(2\) and (2) by \(3\):

\(4a+b=21,\qquad 9a+b=51.\)

Subtracting gives

\(5a=30,\)

so

\(a=6.\)

Then

\(4(6)+b=21,\)

so

\(b=-3.\)

Thus

\(\mathrm p(x)=6x^3-9x^2-3x-6.\)

Since \(x-2\) is a factor, divide by \(x-2\):

\(\mathrm p(x)=(x-2)(6x^2+3x+3).\)

For the quadratic factor, the discriminant is

\(3^2-4(6)(3)=9-72=-63.\)

This is negative, so \(6x^2+3x+3=0\) has no real solutions. Therefore the only real solution of \(\mathrm p(x)=0\) is \(x=2\).

Answer: \(m=6\), \(n=-10\), so \(x-2\) is a factor.

Since \(3x-1\) is a factor, \(x=\frac13\) is a root. Therefore

\(\mathrm p\left(\frac13\right)=0.\)

This gives

\(\frac{m}{27}-\frac{29}{9}+13+n=0. \tag{1}\)

The remainder when divided by \(x-1\) is \(6\), so

\(\mathrm p(1)=6.\)

Thus

\(m-29+39+n=6,\)

or

\(m+n=-4. \tag{2}\)

Solving (1) and (2) gives

\(m=6,\qquad n=-10.\)

Now substitute \(x=2\):

\(\mathrm p(2)=6(2)^3-29(2)^2+39(2)-10.\)

So

\(\mathrm p(2)=48-116+78-10=0.\)

Since \(\mathrm p(2)=0\), \(x-2\) is a factor of \(\mathrm p(x)\).

Answer: \(a=-3\), \(b=2\); remainder \(4\).

Since \(\mathrm p(x)\) is divisible by \(x-2\),

\(\mathrm p(2)=0.\)

So

\(8a+4b+12+4=0,\)

which simplifies to

\(2a+b=-4.\)

Now

\(\mathrm p'(x)=3ax^2+2bx+6.\)

When \(\mathrm p'(x)\) is divided by \(x+1\), the remainder is \(\mathrm p'(-1)\). Hence

\(3a-2b+6=-7,\)

so

\(3a-2b=-13.\)

Using \(b=-4-2a\) in \(3a-2b=-13\),

\(3a-2(-4-2a)=-13.\)

Therefore

\(7a=-21,\)

so \(a=-3\). Then

\(b=-4-2(-3)=2.\)

Using these values,

\(\mathrm p''(x)=6ax+2b=-18x+4.\)

The remainder when \(\mathrm p''(x)\) is divided by \(x\) is \(\mathrm p''(0)\), which is

\(4.\)

Answer: \(a=-13\), \(b=1\); remainder \(0\); \(\mathrm{p}(x)=(x-2)(2x-1)(3x+1)\).

Since \(x-2\) is a factor,

\(\mathrm{p}(2)=0.\)

Therefore

\(6(2)^3+a(2)^2+b(2)+2=0,\)

so

\(48+4a+2b+2=0.\)

This gives

\(2a+b=-25.\)

Also \(\mathrm{p}(0)=2\), so \(\mathrm{p}(1)=-2\mathrm{p}(0)=-4\). Hence

\(6+a+b+2=-4,\)

which gives

\(a+b=-12.\)

Solving

\(2a+b=-25,\qquad a+b=-12,\)

gives

\(a=-13,\qquad b=1.\)

So

\(\mathrm{p}(x)=6x^3-13x^2+x+2.\)

The remainder when dividing by \(2x-1\) is \(\mathrm{p}\left(\frac12\right)\):

\(\mathrm{p}\left(\frac12\right) =6\left(\frac18\right)-13\left(\frac14\right)+\frac12+2 =0.\)

For the factorisation, since \(x-2\) and \(2x-1\) are factors, the remaining factor is \(3x+1\). Therefore

\(\mathrm{p}(x)=(x-2)(2x-1)(3x+1).\)

Answer: \(p(-k)=-12\), \(k=-2\), and \(p(x)=-25\) has no real solutions.

When \(x=-k\), the factor \(x+k\) is zero. Hence

\(p(-k)=(2(-k)-1)(0)-12=-12.\)

By the remainder theorem, the remainder on division by \(x+3\) is \(p(-3)\). So

\(p(-3)=23.\)

Now

\(p(-3)=(-7)(k-3)-12.\)

Therefore

\(-7(k-3)-12=23.\)

So

\(-7k+9=23,\qquad k=-2.\)

Using \(k=-2\),

\(p(x)=(2x-1)(x-2)-12=2x^2-5x-10.\)

The equation \(p(x)=-25\) becomes

\(2x^2-5x-10=-25,\)

or

\(2x^2-5x+15=0.\)

Its discriminant is

\((-5)^2-4(2)(15)=25-120=-95.\)

Since the discriminant is negative, there are no real solutions.

Answer: \(a=10\), \(b=-12\).

By the remainder theorem, the remainder on division by \(x+3\) is \(p(-3)\), and the remainder on division by \(x-2\) is \(p(2)\).

Now

\(p(-3)=-27+9a+9+b=9a+b-18\),

and

\(p(2)=8+4a-6+b=4a+b+2\).

Since \(p(-3)=2p(2)\),

\(9a+b-18=2(4a+b+2)\).

So

\(9a+b-18=8a+2b+4\),

and hence

\(a-b=22.\)

Also, \(p(x)\) is exactly divisible by \(x+1\), so \(p(-1)=0\). Therefore

\(-1+a+3+b=0\),

which gives

\(a+b=-2.\)

Solving

\(a-b=22,\qquad a+b=-2,\)

gives

\(\boxed{a=10,\quad b=-12}.\)

Answer: The remainder is \(60\).

Since \(x+3\) is a factor, \(p(-3)=0\).

\(2(-3)^3+a(-3)^2-24(-3)+b=0.\)

So

\(-54+9a+72+b=0, \qquad 9a+b=-18.\)

The remainder when \(p(x)\) is divided by \(x-2\) is \(p(2)\), so \(p(2)=-15\):

\(2(2)^3+a(2)^2-24(2)+b=-15.\)

Hence

\(16+4a-48+b=-15, \qquad 4a+b=17.\)

Subtracting the two equations gives

\(5a=-35, \qquad a=-7.\)

Then

\(4(-7)+b=17, \qquad b=45.\)

The remainder when \(p(x)\) is divided by \(x+1\) is \(p(-1)\):

\(p(-1)=2(-1)^3-7(-1)^2-24(-1)+45.\)

Therefore

\(p(-1)=-2-7+24+45=60.\)

Answer: \(a=3\), \(b=2\); remainder \(=-\dfrac{35}{27}\).

Since \(p(x)\) has remainder 2 when divided by \(2x+1\), use \(x=-\frac12\):

\(p\left(-\frac12\right)=2.\)

Now

\(2\left(-\frac12\right)^3+5\left(-\frac12\right)^2+4\left(-\frac12\right)+a=a-1.\)

So \(a-1=2\), giving

\(a=3.\)

Since \(q(x)\) is divisible by \(x+2\),

\(q(-2)=0.\)

Thus

\(4(-2)^2+3(3)(-2)+b=0,\)

so

\(16-18+b=0,\qquad b=2.\)

Then

\(r(x)=p(x)-q(x)=2x^3+x^2-5x+1.\)

The remainder when \(r(x)\) is divided by \(3x-2\) is \(r\left(\frac23\right)\):

\(r\left(\frac23\right) =2\left(\frac23\right)^3+\left(\frac23\right)^2-5\left(\frac23\right)+1 =-\frac{35}{27}.\)

Answer: \(a=10\), \(b=43\), \(c=36\); remainder \(21\).

Since \(p(x)\) is divisible by \(x+3\),

\(p(-3)=0.\)

This gives

\(-27a+9b-3c-9=0.\)

Differentiate:

\(p'(x)=3ax^2+2bx+c.\)

Since \(p'(0)=36\),

\(c=36.\)

Differentiate again:

\(p''(x)=6ax+2b.\)

Since \(p''(0)=86\),

\(2b=86,\qquad b=43.\)

Substitute \(b=43\) and \(c=36\) into the equation from \(p(-3)=0\):

\(-27a+9(43)-3(36)-9=0.\)

So

\(-27a+270=0,\qquad a=10.\)

The remainder when dividing by \(2x-1\) is \(p\left(\frac12\right)\). Hence

\(p\left(\frac12\right) =10\left(\frac12\right)^3+43\left(\frac12\right)^2+36\left(\frac12\right)-9 =21.\)

Answer: \(a=12\), \(b=-27\).

By the remainder theorem, the remainder when \(p(x)\) is divided by \(x-3\) is \(p(3)\).

So

\(p(3)=6.\)

Substitute \(x=3\):

\(27+9a+3b-48=6.\)

Hence

\(9a+3b=27,\)

or

\(3a+b=9. \tag{1}\)

Now differentiate:

\(p'(x)=3x^2+2ax+b.\)

Given \(p'(1)=0\),

\(3+2a+b=0.\)

So

\(2a+b=-3. \tag{2}\)

Subtract equation (2) from equation (1):

\(a=12.\)

Then from \(2a+b=-3\),

\(24+b=-3,\)

so

\(b=-27.\)