Tag: factor theorem

Polynomials: Factor and Remainder Theorem

In Algebra, we regularly come across the polynomials. The skill in solving algebraic operations of polynomials is very important in mathematics.

If you need to revise on the basic of algebra, you can look at it here.

The definition

\( {\small P(x) \ = \ a_{n} x^{n} + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + … + a_{0} }\)

\(\\[5pt] {\small n \ }\) is a non-negative integer. It is the degree or the order of the polynomial.

\(\\[5pt] {\small x \ }\) is the variable of the polynomial.

\(\\[5pt] {\small a_{n}, a_{n-1}, a_{n-2}, … , a_{0} \ }\) are the coefficients of each term in \({ \small x \ }\) from the highest power to the lowest power.

Here are some of the examples:

1. \(\\[5pt]\) Linear polynomial or polynomial of degree 1
\( \hspace{1em} {\small P(x) \ = \ ax \ + \ b }\)
e.g.: \({\small P(x) \ = \ 2x \ + \ 3 }\)

2. \(\\[5pt]\) Quadratic polynomial or polynomial of degree 2
\( \hspace{1em} {\small P(x) \ = \ ax^{2} + bx + c }\)
e.g.: \({\small P(x) \ = \ 3x^{2} + 2x + 3 }\)

3. \(\\[5pt]\) Cubic polynomial or polynomial of degree 3
\( \hspace{1em} {\small P(x) \ = \ ax^{3} + bx^{2} + cx + d }\)
e.g.: \({\small P(x) \ = \ 3x^{3} + 2x^2 + 3x + 5 }\)

4. \(\\[5pt]\) Quartic polynomial or polynomial of degree 4
\( \hspace{1em} {\small P(x) \ = \ ax^{4} + bx^{3} + cx^{2} + dx + e }\)
e.g.: \({\small P(x) \ = \ 5x^{4} + 2x^{3} + 3x^2 + x + 6 }\)

The addition and subtraction of polynomials

Collect all the like terms and simply the expression by adding or subtracting the like terms.

e.g.:

\( \hspace{1em} {\small P(x) \ = \ 2x^{2} + 3x + 4 }\)
\( \hspace{1em} {\small Q(x) \ = \ x^{2} \ – \ 2x + 3 }\)

\( \hspace{1em} {\small P(x) \ + \ Q(x) }\)
\( \hspace{1.5em} {\small = \ (2x^{2} + 3x + 4) + (x^{2} \ – \ 2x + 3) }\)
\( \hspace{1.5em} {\small = \ (2x^{2} + x^{2}) + (3x \ – \ 2x) + (4 + 3) }\)
\( \hspace{1.5em} {\small = \ 3x^{2} + x + 7 }\)

\( \hspace{1em} {\small P(x) \ – \ Q(x) }\)
\( \hspace{1.5em} {\small = \ (2x^{2} + 3x + 4) \ – \ (x^{2} \ – \ 2x + 3) }\)
\( \hspace{1.5em} {\small = \ (2x^{2} \ – \ x^{2}) + (3x \ + \ 2x) + (4 \ – \ 3) }\)
\( \hspace{1.5em} {\small = \ x^{2} + 5x + 1 }\)

The multiplication of polynomials

Use the distributive law by multiplying each term.

e.g.:

\( \hspace{1em} {\small P(x) \ = \ 2x^{2} + 3x + 4 }\)
\( \hspace{1em} {\small Q(x) \ = \ x^{2} \ – \ 2x + 3 }\)

\( \hspace{1em} {\small P(x) \ \times \ Q(x) }\)
\( \hspace{1.5em} {\small = \ 2x^{4} + (3x^{3} \ – \ 4x^{3}) + (6x^{2} \ – \ 6x^{2} + 4x^{2}) }\)
\(\\[8pt] \hspace{2.5em} {\small + \ (9x \ – \ 8x) + 12 }\)
\( \hspace{1.5em} {\small = \ 2x^{4} \ – \ x^{3} + 4x^{2} + x + 12 }\)

The division of polynomials

We’ll use a simple analogy of long division of a number to illustrate the polynomial division.

Just as an example, what is the result of 7 divided by 3?

So, \(\hspace{1em} {\large\frac{7}{3}} \ = \ 2 \ + \ {\large\frac{1}{3}} \)

Or, we can conveniently write the above result as,

\(\hspace{1.5em} 7 \ = \ 2 \times 3 \ + \ 1 \ \)

Then, in terms of polynomial division,

\( \hspace{1em} {\large\frac{P(x)}{D(x)}} \ = \ Q(x) \ + {\large\frac{R(x)}{D(x)}} \)

or in its general form:

\( \hspace{1em} P(x) \ = \ D(x) \times Q(x) + R(x) \)

with,
P(x) is the polynomial to be divided with or the dividend,
D(x) is the divisor,
Q(x) is the quotient and
R(x) is the remainder.

The process of finding the quotient Q(x) and the remainder R(x) can be done similarly with polynomial long division.

Here is an example,

\(\hspace{1em} {\Large\frac{x^{2} \ – \ 4x \ – \ 3}{x \ + \ 1}} \ = \ ? \)

Step 1. Divide the highest power of the polynomial P(x) with the highest power of the divisor D(x).

Step 2. Multiply the result we get in the quotient Q(x) from Step 1 with the divisor D(x).

Step 3. Subtract the polynomial P(x) from the result in Step 2.

Step 4. Repeat the process from Step 1 to the next available highest power of polynomial P(x) until we are left with only the remainder.

Now, one important thing to remember is that the degree of the remainder R(x) is at most one fewer than the order of the divisor D(X).

For example, if we have a linear divisor \({\small \ ax \ + \ b \ }\) then the remainder is a constant.

If we have a quadratic divisor \({\small \ ax^{2} \ + \ bx \ + \ c \ }\) then the remainder is a linear polynomial and so on.

By knowing that, we’ll know when to stop the process of multiplying back the quotient.

Factor Theorem

If \({\small \ (ax \ + \ b) \ }\) is a linear factor of a polynomial P(x) then \({\small \ P({\large\frac{-b}{a}}) \ = \ 0 }\).

e.g.: Show that \({\small \ (x \ – \ 2) \ }\) is a linear factor of a polynomial \({\small P(x) \ = \ x^{4} \ – \ 5x^{2} + 2x }\).

Solution: If \({\small \ (x \ – \ 2) \ }\) is a linear factor, then according to the factor theorem \({\small \ P(2) \ }\) must be equal to zero.

We’ll try by substituting the value of \({\small \ x \ = \ 2 }\),

\(\hspace{1.5em} {\small P(2) \ = \ ({2})^{4} \ – \ 5({2})^{2} + 2(2) }\)
\(\hspace{3.2em} {\small \ = \ 16 \ – \ 20 + 4 }\)
\(\hspace{3.2em} {\small \ = \ 0 }\)

Therefore, \({\small \ (x \ – \ 2) \ }\) is a linear factor of \({\small \ P(x) \ = \ x^{4} \ – \ 5x^{2} + 2x }\).

Another great use of factor theorem is when we want to factorise a higher order polynomial or the roots of a higher order polynomial equation.

We can use the constant term of the higher order polynomial and quickly substituting the value of each factors of the constant term.

If any of the results equals to zero then by factor theorem we can find its linear factor(s).

The other factors can then be found by applying polynomial long division.

You can study some of the examples below to fully understand the concept.

Remainder Theorem

If a polynomial P(x) is divided by a linear divisor \({\small \ (ax \ + \ b) \ }\) then \({\small \ P({\large\frac{-b}{a}}) \ }\) is the remainder.

e.g.: Find the remainder when a polynomial \({\small \ P(x) \ = \ x^{4} \ – \ 2x^{3} + 3x^{2} \ – \ 8 \ }\) is divided by \({\small \ (x \ – \ 1) }\).

Solution: According to the remainder theorem, the remainder is \({\small \ P(1) \ }\),

\(\hspace{1.5em} {\small P(1) \ = \ (1)^{4} \ – \ 2(1)^{3} + 3(1)^{2} \ – \ 8 }\)
\(\hspace{3.2em} {\small \ = \ 1 \ – \ 2 + 3 \ – \ 8 }\)
\(\hspace{3.2em} {\small \ = \ -6 }\)

Therefore, the remainder of \({\small \ P(x) \ = \ x^{4} \ – \ 2x^{3} + 3x^{2} \ – \ 8 \ }\) is \({\small \ -6 \ }\) when it is divided by \({\small \ (x \ – \ 1) }\).

Try some of the examples below and if you need any help, just look at the solution I have written. Cheers ! =) .
\(\\[1pt]\)

EXAMPLE:

\({\small 1.\enspace}\) Find the remainder of \({\small \ P(x) \ = \ 2x^{3} + 4x^{2} \ – \ 6x + 7 \ }\) when it is divided by \({\small \ (2x \ – \ 1) }\).

Show solution:

We can use polynomial long division to find the remainder.
\(\\[1pt]\)

\(\\[1pt]\)
Or, we can use the remainder theorem to quickly find the answer to the question.
\(\\[1pt]\)
First, we find the value of \({\small x \ }\) so that the linear divisor \({\small \ (2x \ – \ 1) \ }\) is zero.
\(\\[1pt]\)
\( \\[10pt]\hspace{3em} {\small \ (2x \ – \ 1) \ = \ 0 }\)
\( \\[10pt]\hspace{5em} {\small \ 2x \ = \ 1 }\)
\( \hspace{5.4em} {\small \ x \ = \ {\large \frac{1}{2} }}\)
\(\\[1pt]\)
\(\\[15pt]\) Then, the remainder is \({\small \ P( \frac{1}{2}) }\),
\(\\[12pt]\hspace{1.5em} {\small P( \frac{1}{2}) \ = \ 2{( \frac{1}{2})}^{3} + 4{( \frac{1}{2})}^{2} \ – \ 6( \frac{1}{2}) + 7 }\)
\(\\[12pt]\hspace{3.4em} {\small \ = \ \frac{1}{4} + 1 \ – \ 3 + 7 }\)
\(\hspace{3.4em} {\small \ = \ \frac{21}{4} }\)

\(\\[1pt]\)
\({\small 2.\enspace}\) Find the value of \({\small \ k \ }\) if \({\small \ (x \ – \ 1) \ }\) is a factor of \({\small \ {k}^{2}{x}^{4} \ – \ 3k{x}^{2} + 2 }\).

Show solution:

\(\\[10pt]\) By using the factor theorem,
Since \({\small \ (x \ – \ 1) \ }\) is a factor of \({\small \ {k}^{2}x^{4} \ – \ 3k{x}^{2} + 2 }\), then
\(\\[1pt]\)
\(\\[10pt]\hspace{3em} {\small \ {k}^{2}{(1)}^{4} \ – \ 3k{(1)}^{2} + 2 \ = \ 0}\)
\(\\[10pt]\hspace{6em} {\small \ {k}^{2} \ – \ 3k + 2 \ = \ 0}\)
\(\\[10pt]\hspace{5.1em} {\small \ (k \ – \ 2)(k \ – \ 1) \ = \ 0}\)
\(\hspace{3em} {\small \ {k}_{1} \ = \ 2 \ \ \ \textrm{or} \ \ \ {k}_{2} \ = \ 1}\)

\(\\[1pt]\)
\({\small 3.\enspace}\) This is an example of the aplication of the factor theorem. We would try to find the roots of higher order polynomials.
\(\\[1pt]\)
Factorise \({\small \ 2{?}^{3} + 9{?}^{2} + 7{?} \ − \ 6 \ }\) completely and hence find all the roots of the polynomial.

Show solution:

Since this is a cubic polynomial or a higher order polynomial, we can’t use our knowledge from solving the roots of quadratic equations.
\(\\[1pt]\)
So, we can start by taking the constant term in the polynomial, -6.
\(\\[1pt]\)
Next, we find all the possible factors of -6: \({\small \ \pm 1, \pm 2, \pm 3, \pm 6 }\).
\(\\[1pt]\)
And then, we try to find the value of the polynomials at those factors.
\(\\[1pt]\)
By factor theorem we know if we get a zero value at any of those factors then it is a factor of the polynomial.
\(\\[1pt]\)
\(\\[12pt]\) Let’s start from \({\small \ x \ = \ 1 }\),
\(\\[10pt]\hspace{1.5em} {\small P( 1 ) \ = \ 2{(1)}^{3} + 9{(1)}^{2} + 7{(1)} \ − \ 6 }\)
\(\\[10pt]\hspace{3.7em} {\small = \ 2 + 9 + 7 \ − \ 6 }\)
\(\\[12pt]\hspace{3.7em} {\small = \ 12 }\)
Since \( \ {\small P( 1 ) \ \neq \ 0 \ }\) then \({\small \ (x \ – \ 1) \ }\) is not a factor.
\(\\[1pt]\)
We’ll proceed with the other factors of -6 till we find the zeros.
\(\\[1pt]\)
\(\\[12pt]\) Use \({\small \ x \ = \ -1 }\),
\(\\[10pt]\hspace{1.2em} {\small P( -1 ) \ = \ 2{(-1)}^{3} + 9{(-1)}^{2} + 7{(-1)} \ − \ 6 }\)
\(\\[10pt]\hspace{4em} {\small = \ -2 + 9 \ – \ 7 \ − \ 6 }\)
\(\\[12pt]\hspace{4em} {\small = \ -6 }\)
Since \( \ {\small P( -1 ) \ \neq \ 0 \ }\) then \({\small \ (x + 1) \ }\) is not a factor.
\(\\[1pt]\)
\(\\[12pt]\) Use \({\small \ x \ = \ 2 }\),
\(\\[10pt]\hspace{1.5em} {\small P( 2 ) \ = \ 2{(2)}^{3} + 9{(2)}^{2} + 7{(2)} \ − \ 6 }\)
\(\\[10pt]\hspace{3.7em} {\small = \ 16 + 36 + 14 \ − \ 6 }\)
\(\\[12pt]\hspace{3.7em} {\small = \ 60 }\)
Since \( \ {\small P( 2 ) \ \neq \ 0 \ }\) then \({\small \ (x \ – \ 2) \ }\) is not a factor.
\(\\[1pt]\)
\(\\[12pt]\) Use \({\small \ x \ = \ -2 }\),
\(\\[10pt]\hspace{1.2em} {\small P( -2 ) \ = \ 2{(-2)}^{3} + 9{(-2)}^{2} + 7{(-2)} \ − \ 6 }\)
\(\\[10pt]\hspace{4em} {\small = \ -16 + 36 \ – \ 14 \ − \ 6 }\)
\(\\[12pt]\hspace{4em} {\small = \ 0 }\)
Since \( \ {\small P( -2 ) \ =\ 0 \ }\) then \({\small \ (x + 2) \ }\) is a factor.
\(\\[1pt]\)
Once you have found one of the factors, you can continue the trial and error method with the rest of the factors of the constant term.
\(\\[1pt]\)
Or, you can find the other factors by polynomial long division when the remainder has been reduced to a quadratic polynomial.
\(\\[1pt]\)
And then proceed by factorizing the remainder using the techniques we have learned before in solving quadratic equations.
\(\\[1pt]\)
With one of the linear factor is now known \({\small \ (x + 2) }\), i’ll proceed with polynomial long division and solve the remaining roots.
\(\\[1pt]\)

\(\\[1pt]\)
\( {\small 2{?}^{3} + 9{?}^{2} + 7{?} \ − \ 6 \ = \ (x + 2)(2{x}^2 + 5x \ – \ 3) }\)
\(\\[1pt]\)
Then we can factorise the quadratic factor \({\small \ (2{x}^2 + 5x \ – \ 3) \ }\) into its linear factors,
\(\\[1pt]\)
\(\\[10pt]\hspace{2.8em} {\small \ 2{x}^2 + 5x \ – \ 3 \ = \ (2x \ – \ 1)(x + 3)}\)
\(\\[10pt]{\small \ 2{?}^{3} + 9{?}^{2} + 7{?} \ − \ 6 \ = \ (x + 2)(2x \ – \ 1)(x + 3)}\)
\(\\[1pt]\)
\(\\[12pt]\) And the roots are:
\(\\[10pt]\hspace{3.7em} {\small \ 2{?}^{3} + 9{?}^{2} + 7{?} \ − \ 6 \ = \ 0}\)
\(\\[10pt]\hspace{3em} {\small \ (x + 2)(2x \ – \ 1)(x + 3) \ = \ 0}\)
\(\hspace{1em} {\small \ {x}_{1} \ = \ -2 \ \ \ \textrm{or} \ \ \ {x}_{2} \ = \ {\large\frac{1}{2}} \ \ \ \textrm{or} \ \ \ {x}_{3} \ = \ -3}\)

\(\\[1pt]\)
\({\small 4.\enspace}\) Given that a polynomial \({\small \ p(x) \ = \ {?}^{3} + a{?}^{2} + b{?} \ − \ 3 \ }\) when divided by \({\small \ (x + 1) \ }\) and \({\small \ (x \ – \ 1) }\), the remainders are -9 and 1 respectively. Find the values of a and b.

Show solution:

Due to the remainder theorem, when the polynomial \({\small \ p(x) \ = \ {?}^{3} + a{?}^{2} + b{?} \ − \ 3 \ }\) is divided by \({\small \ (x + 1) }\), the remainder is \({\small \ p(-1) }\).
\(\\[1pt]\)
\(\\[10pt]\hspace{6em} {\small p(-1) \ = \ -9}\)
\(\\[10pt]\hspace{1.2em} {\small {(-1)}^{3} + a{(-1)}^{2} + b{(-1)} \ − \ 3 \ = \ -9 }\)
\(\\[10pt]\hspace{3em} {\small -1 + a \ – \ b \ – \ 3 \ = \ -9 }\)
\(\\[10pt]\hspace{6.2em} a \ – \ b \ = \ -5 \)
\(\\[1pt]\)
Again, when the polynomial \({\small \ p(x) \ = \ {?}^{3} + a{?}^{2} + b{?} \ − \ 3 \ }\) is divided by \({\small \ (x \ – \ 1) }\), the remainder is \({\small \ p(1) }\).
\(\\[1pt]\)
\(\\[10pt]\hspace{6.2em} {\small p(1) \ = \ 1}\)
\(\\[10pt]\hspace{1.2em} {\small {(1)}^{3} + a{(1)}^{2} + b{(1)} \ − \ 3 \ = \ 1 }\)
\(\\[10pt]\hspace{3em} {\small 1 + a \ + \ b \ – \ 3 \ = \ 1 }\)
\(\\[17pt]\hspace{5.8em} a + b \ = \ 3 \)
Solving for both equations in a and b simultaneously by substitution or elimination method,
\(\\[1pt]\)
\(\\[7pt]\hspace{2em} {\small a \ = \ -1 }\)
\(\\[17pt]\hspace{2em} {\small b \ = \ 4 }\)
Thus, the polynomial is \({\small \ p(x) \ = \ {?}^{3} \ – \ {?}^{2} + 4{?} \ − \ 3 }\).

\(\\[1pt]\)
\({\small 5.\enspace}\) Continuing from the same question in example 4, find the reminder when the polynomial \({\small \ p(x) \ }\) is divided by \({\small \ ({x}^{2} \ – \ 1) }\).

Show solution:

We can use polynomial long division to find the remainder.
\(\\[1pt]\)

\(\\[1pt]\)
The remainder is \({\small \ \ 5{?} \ − \ 4 }\).
\(\\[1pt]\)
Alternatively, we can use the remainder theorem. Since we have a quadratic divisor \({\small \ ({x}^{2} \ – \ 1)}\), the remainder is a linear polynomial.
\(\\[1pt]\)
Remember that the order of the remainder R(x) is at most one fewer than the order of the divisor D(X).
\(\\[1pt]\)
Let: \(\\[12pt]\hspace{1em}{\small R(x) \ = \ px + q \ }\)
with p and q are the two unknown constants we’re about to solve in a short while.
\(\\[1pt]\)
We can write in the general form of polynomial division,
\(\\[1pt]\)
\(\\[10pt]\hspace{3.8em} {\small P(x) \ = \ D(x) \times Q(x) + R(x) }\)
\(\\[20pt]{\small {?}^{3} – {?}^{2} + 4{?} − 3 = ({x}^{2} – 1) \times Q(x) + px + q }\)
\(\\[15pt]\) Also, we can factorise the divisor \({\small \ ({x}^{2} – 1) }\),
\(\\[15pt]{\small {?}^{3} – {?}^{2} + 4{?} − 3 = (x + 1)(x – 1) \times Q(x) + px + q }\)
\(\\[12pt]\) At \({\small \ x \ = \ -1 }\),
\(\\[12pt]\hspace{2em} \small {D(x) \times Q(x) \ = \ 0 \ }\) since \({\small \ ({x}^{2} – 1) \ = \ 0 }\),
\(\\[12pt]\) Only the remainder term remains.
\(\\[12pt]\) Then, by using the remainder theorem,
\(\\[10pt]\hspace{2em} {\small p( -1 ) \ = \ p(-1) + q }\)
\(\\[10pt]\hspace{3.1em} {\small -9 \ = \ -p + q}\)
\(\\[1pt]\)
\(\\[12pt]\) At \({\small \ x \ = \ 1 }\),
\(\\[12pt]\hspace{2em} \small {D(x) \times Q(x) \ = \ 0 \ }\) since \({\small \ ({x}^{2} – 1) \ = \ 0 }\),
\(\\[12pt]\) Only the remainder term remains.
\(\\[12pt]\) Then, by using the remainder theorem,
\(\\[10pt]\hspace{2em} {\small p( 1 ) \ = \ p(1) + q }\)
\(\\[10pt]\hspace{3.1em} {\small 1 \ = \ p + q}\)
\(\\[1pt]\)
Solving for both equations in p and q simultaneously by substitution or elimination method,
\(\\[1pt]\)
\(\\[7pt]\hspace{2em} {\small p \ = \ 5 }\)
\(\\[17pt]\hspace{2em} {\small q \ = \ -4 }\)
Thus, we arrive at the same answer with the remainder is \({\small \ \ 5{?} \ − \ 4 }\).

\(\\[1pt]\)
\({\small 6.\enspace}\) 9709/21/O/N/19 – Paper 21 March 2019 Pure Maths 2 No 4
\(\\[1pt]\)
The polynomial p(x) is defined by:
\(\\[1pt]\)
\(\hspace{3em} {\small p(x) \ = \ a{x}^{3} + a{x}^{2} \ − \ 15x \ − \ 18}\),
\(\\[1pt]\)
where a is a constant. It is given that (x − 2) is a factor of p(x).
\({\small (\textrm{i}).\hspace{0.7em}}\) Find the value of a.
\({\small (\textrm{ii}).\hspace{0.6em}}\) Using this value of a, factorise p(x) completely.
\({\small (\textrm{iii}).\hspace{0.5em}}\) Hence solve the equation \({\small \ p(e^{\sqrt{y}}) \ = \ 0}\), giving the answer correct to 2 significant figures.

Show solution:

\(\\[12pt]\hspace{1.2em} {\small (\textrm{i}).\hspace{0.7em}}\) Since (x − 2) is a factor of p(x),
\(\\[10pt]\hspace{6.7em} {\small \ p(2) \ = \ 0}\)
\(\\[10pt] {\small \ a{(2)}^{3} + a{(2)}^{2} \ − \ 15(2) \ − \ 18 \ = \ 0}\)
\(\\[10pt]\hspace{1.8em} {\small \ 8a + 4a \ – \ 30 \ – \ 18 \ = \ 0}\)
\(\\[10pt]\hspace{7em} {\small \ 12a \ = \ 48}\)
\(\\[15pt] \hspace{7.9em} {\small \ a \ = \ 4 }\)
Therefore, \( {\small p(x) \ = \ 4{x}^{3} + 4{x}^{2} \ − \ 15x \ − \ 18 }\).
\(\\[1pt]\)
\(\hspace{1.2em} {\small (\textrm{ii}).\hspace{0.6em}}\) \({\small (x \ – \ 2) \ }\) is one of the factors of p(x).
\(\\[1pt]\)
We can try to find the other factors of p(x) by using the factors of the constant term, -18: \({\small \ \pm 1, -2, \pm 3, \pm 6, \pm 9, \pm 18}\).
\(\\[1pt]\)
If we get a zero value when substituting the (x) variable of p(x) at any of those factors then it is a factor of the polynomial.
\(\\[1pt]\)
Or, alternatively we can use polynomial long division to find the other quadratic factor and then completely factorised it into its linear factors.
\(\\[1pt]\)

\(\\[1pt]\)
\(\\[12pt]\) Then, we can write :
\( {\small 4{?}^{3} + 4{?}^{2} – 15{?} -18 = (x – 2)(4{x}^2 + 12x + 9) }\)
\(\\[1pt]\)
The quadratic factor can be factorised further as linear factors,
\(\\[1pt]\)
\(\\[10pt]\hspace{2.8em} {\small \ 4{x}^2 + 12x + 9 \ = \ {(2x + 3)}^{2} }\)
\(\\[1pt]\)
Therefore, the polynomial p(x) can be completely factorised as below,
\(\\[1pt]\)
\(\hspace{1.2em} {\small 4{?}^{3} + 4{?}^{2} – 15{?} -18 = (x – 2){(2x + 3)}^{2} }\)
\(\\[1pt]\)
\(\\[15pt] \hspace{1.2em} {\small (\textrm{iii}).\hspace{0.5em}}\) Let \({\small \ e^{ \sqrt{y} } \ = \ k}\)
\(\\[10pt] \hspace{4em} {\small \ p(k) \ = \ 0}\)
\(\\[10pt] \hspace{2em} {\small 4{k}^{3} + 4{k}^{2} \ – \ 15{k} \ – \ 18 = 0 }\)
\(\\[10pt] \hspace{3.7em} {\small (k \ – \ 2){(2k + 3)}^{2} \ = \ 0 }\)
\(\\[10pt] \hspace{2em} {\small \ (k \ – \ 2) \ = \ 0 \ \ \ \textrm{or} \ \ \ {(2k + 3)}^{2} \ = \ 0}\)
\(\\[17pt] \hspace{3.7em} {\small \ {k}_{1} \ = \ 2 \ \ \ \textrm{or} \ \ \ {k}_{2} \ = \ -\frac{3}{2}}\)
\( {\small {k}_{2} \ = \ -\frac{3}{2} \ }\) is rejected since \({\small \ e^{ \sqrt{y} } \ = \ k \ > \ 0 }\).
\(\\[1pt]\)
Remember that exponential function is always greater than zero.
\(\\[1pt]\)
\(\\[10pt]\) Therefore,
\(\\[15pt] \hspace{3.8em} {\small \ k \ = \ 2 }\)
\(\\[15pt] \hspace{3em} {\small \ e^{ \sqrt{y} } \ = \ 2 }\)
\(\\[15pt] \hspace{3.1em} {\small \ \sqrt{y} \ = \ \ln{2}}\)
\(\\[15pt]\) Squaring both sides of the equation,
\(\\[12pt] \hspace{3em} {\small \ y \ = \ {(\ln{2})}^{2} }\)
\( \hspace{3em} {\small \ y \ \approx \ 0.48 }\)

\(\\[1pt]\)
\({\small 7.\enspace}\) 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 1
\(\\[1pt]\)
Find the quotient and remainder when \(6x^4 + x^3 − x^2 + 5x − 6\) is divided by \( 2x^2 − x + 1 \).

Show solution:

Check out my solution here:
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https://www.instagram.com/p/CalWVL8pq_f/
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\(\\[1pt]\)
\({\small 8.\enspace}\) 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 5(a)
\(\\[1pt]\)
Find the quotient and remainder when \(2x^3 − x^2 + 6x + 3\) is divided by \(x^2 + 3\).

Show solution:

Check out my solution here:
\(\\[1pt]\)
https://www.instagram.com/p/Cbgi9MVPbxl/
\(\\[1pt]\)
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\({\small 9.\enspace}\) 9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 2
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The polynomial \(ax^3 + 5x^2 − 4x + b\), where \(a\) and \(b\) are constants, is denoted by \(p(x)\). It is given that
\( (x + 2) \) is a factor of \(p(x)\) and that when \(p(x)\) is divided by \( (x + 1) \) the remainder is 2.
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Find the values of \(a\) and \(b\).

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\({\small 10.\enspace}\) 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 8a
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Find the quotient and remainder when \({\small 8x^3 + 4x^2 + 2x + 7}\) is divided by \({\small 4x^2 + 1}\).

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PRACTICE MORE WITH THESE QUESTIONS BELOW!

\({\small 1.\enspace}\) The polynomial p(x) is defined by

\( \hspace{3em} {\small \ p(x) \ = \ 6{x}^{3} + a{x}^{2} + 9{x} + b }\),

where a and b are constants. It is given that \({\small \ (x \ – \ 2) \ }\) and \({\small \ (2x + 1) }\) are factors of p(x). Find the values of a and b.

\({\small 2. \enspace}\) The polynomial p(x) is defined by

\( \hspace{3em} {\small \ p(x) \ = \ 6{x}^{3} + a{x}^{2} \ – \ 4{x} \ – \ 3 }\),

where a is a constant. It is given that \({\small \ (x + 3) \ }\) is a factor of p(x).
\({\small\hspace{1.2em} (\textrm{a}).\hspace{0.8em}}\) Find the value of a.
\({\small\hspace{1.2em} (\textrm{b}).\hspace{0.8em}}\) Using this value of a, factorise p(x) completely.

\({\small 3. \enspace}\) Find the quotient and remainder when \({\small \ 6{x}^{4} + {x}^{3} \ − \ {x}^{2} + 5x \ − \ 6 \ }\) is divided by \( {\small \ (2{x}^{2} \ − \ x + 1) }\).

\({\small 4. \enspace}\) The polynomial f(x) is defined by

\( \hspace{3em} {\small \ f(x) \ = \ {x}^{4} \ – \ 3{x}^{3} + 5{x}^{2} \ – \ 6{x} + 11 }\),

Find the quotient and remainder when f(x) is divided by \({\small \ ({x}^2 + 2) }\).

\({\small 5. \enspace}\) The polynomial \({\small \ 6{x}^{3} + a{x}^{2} + b{x} \ − \ 2 }\), where a and b are constants, is denoted by p(x). It is given that \({\small \ (2x + 1) }\) is a factor of p(x) and that when p(x) is divided by \({\small \ (x + 2) \ }\) the remainder is −24. Find the values of a and b.

\({\small 6. \enspace}\) The polynomial \({\small \ {x}^{4} + 3{x}^{3} + ax + b }\), where a and b are constants, is denoted by p(x). When p(x) is divided by \({\small \ {x}^{2} + x \ − \ 1 \ }\) the remainder is \({\small \ 2x + 3}\). Find the values of a and b.

\({\small 7. \enspace}\) The cubic polynomial f(x) is defined by \({\small \ f(x) \ = \ {x}^{3} + a{x}^{2} + 14x + a + 1 }\), where a is a constant. It is given that \({\small \ (x + 2) \ }\) is a factor of f(x). Use the factor theorem to find the value of a and hence factorise f(x) completely.

\({\small 8. \enspace}\) \({\small (\textrm{i}).\hspace{0.7em}}\) Find the quotient when \({\small \ {x}^{4} \ − \ 2{x}^{3} + 8{x}^{2} \ − \ 12x + 13 \ }\) is divided by \({\small \ {x}^{2} + 6 \ }\) and show that the remainder is 1.
\({\small\hspace{1.3em}(\textrm{ii}).\hspace{0.7em}}\) Show that the equation \({\small \ {x}^{4} \ − \ 2{x}^{3} + 8{x}^{2} \ − \ 12x + 12 \ = \ 0 \ }\) has no real roots.

\({\small 9. \enspace}\) The polynomial \({\small \ {x}^{4} + {x}^{3} + ax + b}\), where a and b are constants, is divisible by \({\small \ {x}^{2} \ − \ x + 1}\). Find the values of a and b.

\({\small 10.\enspace}\) The polynomial p(x) is defined by \({\small \ p(x) \ = \ 4{x}^{3} + 4{x}^{2} \ − \ 29x \ − \ 15 }\).
\({\small\hspace{2.8em}(\textrm{i}).\hspace{0.7em}}\) Use the factor theorem to show that \({\small \ x + 3 \ }\) is a factor of p(x).
\({\small\hspace{2.8em}(\textrm{ii}).\hspace{0.7em}}\) Factorise p(x) completely.

As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .