DBE CAPS Mathematics
Confirm the Grade 12 Polynomials requirement: cubics and theorem application through degree 3.
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Grade 12 CAPS · Practice in layers
Polynomials:
Polynomials:
Mastery Bank
18 original questions. Start at the top, show your own working, then reveal the method. The bank builds from reading a divisor to solving full cubics.
A
Warm-up · 4 questions
Read the algebra before you calculate
These skills stop sign and missing-term errors from contaminating the theorem work.
Set A Foundations
1 mark For \(p(x)=-4x^3+2x-7\), state the degree and leading coefficient.
The largest exponent is \(3\), so the degree is 3. The leading coefficient is \(-4\).
1 mark Rewrite \(2x^3-5x+1\) with every power shown.
2x³ + 0x² − 5x + 1
The missing \(x^2\) term has coefficient \(0\).
1 mark If the divisor is \(x+4\), what value of \(x\) do you use?
x + 4 = 0 ⇒ x = −4
Use \(-4\).
1 mark If the divisor is \(3x+2\), what value of \(x\) do you use?
3x + 2 = 0 ⇒ x = −2/3
Use \(-\frac23\).
B
Remainders · 5 questions
Read the divisor, then substitute
Write the substitution value before evaluating. It is the mark-saving habit.
Set B Remainder Theorem
2 marks Find the remainder when \(x^3-2x^2+3x-4\) is divided by \(x+2\).
\(x+2=0\Rightarrow x=-2\).
p(−2) = −8 − 8 − 6 − 4 = −26
Remainder = \(-26\).
2 marks Find the remainder when \(4x^3-3x+4\) is divided by \(2x-1\).
\(2x-1=0\Rightarrow x=\frac12\).
p(1/2) = 4(1/8) − 3(1/2) + 4 = 3
Remainder = \(3\).
2 marks Find the remainder when \(x^2+5x-1\) is divided by \(2x-3\).
\(2x-3=0\Rightarrow x=\frac32\).
p(3/2) = 9/4 + 15/2 − 1 = 35/4
Remainder = \(\frac{35}{4}\).
2 marks Express the remainder when \(p(x)=x^3-3x+4\) is divided by \(x+m\), in terms of \(m\).
\(x+m=0\Rightarrow x=-m\).
p(−m) = −m³ + 3m + 4
Remainder = \(-m^3+3m+4\).
2 marks Find the remainder when \(2x^3+x^2-7x+4\) is divided by \(2x+1\).
\(2x+1=0\Rightarrow x=-\frac12\).
2(−1/2)³ + (−1/2)² − 7(−1/2) + 4 = 15/2
Remainder = \(\frac{15}{2}\).
C
Factors & parameters · 5 questions
Turn zero into a useful factor
Every zero remainder should become a conclusion and then a new step.
Set C Factor Theorem
3 marks Show that \(x+1\) is a factor of \(x^3+x^2-4x-4\), then factorise fully.
p(−1) = −1 + 1 + 4 − 4 = 0
So \(x+1\) is a factor.
x³ + x² − 4x − 4 = (x+1)(x²−4) = (x+1)(x−2)(x+2)
2 marks Is \(x-2\) a factor of \(2x^3-3x^2-8x-3\)?
p(2) = 16 − 12 − 16 − 3 = −15
The remainder is nonzero, so \(x-2\) is not a factor.
2 marks Given that \(x-2\) is a factor of \(x^3+kx^2-5x-6\), find \(k\).
p(2) = 8 + 4k − 10 − 6 = 0
4k − 8 = 0 ⇒ k = 2
3 marks \(q(x)=2x^3+px^2-7x+4\) leaves remainder \(5\) when divided by \(2x+1\). Find \(p\).
Use \(x=-\frac12\):
−1/4 + p/4 + 7/2 + 4 = 5
(p + 29)/4 = 5 ⇒ p = −9
4 marks \(P(x)=x^3+px^2-7x+q\). If \(x+1\) is a factor and the remainder on division by \(x-2\) is \(-9\), find \(p\) and \(q\).
Factor condition: \(P(-1)=0\Rightarrow p+q+6=0\).
Remainder condition: \(P(2)=-9\Rightarrow 4p+q=-3\).
q = −p − 6; 4p − p − 6 = −3 ⇒ p = 1, q = −7
Check: \(P(x)=x^3+x^2-7x-7=(x+1)(x^2-7)\).
D
Cubic equations · 4 questions
Factor completely, then solve completely
These questions test whether you can carry the factor theorem all the way to roots.
Set D Cubics
4 marks Solve \(x^3-4x^2+x+6=0\).
\(p(-1)=0\), so \(x+1\) is a factor.
x³ − 4x² + x + 6 = (x+1)(x²−5x+6)
(x+1)(x−2)(x−3) = 0
\(x=-1, 2, 3\).
4 marks Solve \(2x^3+x^2-8x-4=0\).
\(p(2)=0\), so \(x-2\) is a factor.
2x³ + x² − 8x − 4 = (x−2)(2x²+5x+2)
(x−2)(2x+1)(x+2) = 0
\(x=2, -\frac12, -2\).
3 marks Solve \(x^3-3x^2+4=0\), stating the repeated root.
x³ − 3x² + 4 = (x−2)²(x+1)
\(x=2\) (double root), \(x=-1\).
4 marks Solve \(x^3-5x^2+6x-2=0\).
\(p(1)=0\), so \(x-1\) is a factor.
x³ − 5x² + 6x − 2 = (x−1)(x²−4x+2)
x = (4 ± √(16−8))/2 = 2 ± √2
\(x=1, 2+\sqrt2, 2-\sqrt2\).
Finished all four sets? You have touched every core Grade 12 CAPS Polynomials move. Now test it without reveal answers.Open 12-mark test
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