Practice Questions

P(x)=ax+b
P(0)= a x0 +b = >b=3
P(1)= ax1 +b
a+b=0
a+3=0, a=-3
P(x)=-3x+3

x² - 4x+ 4
product = 4
Sum=-4
1x4=4
-2 x -2 =4
(x-2)(x-2)

<h3 style="color: rgb(255,255,255);">Polynomial Factorization Techniques</h3>
<p style="color: rgb(255,255,255);">To find the factors of a cubic polynomial like $P(x) = x^3+2x^2-5x-6$, several methods can be employed, which are commonly tested in competitive exams.</p>

<h4 style="color: rgb(255,255,255);">1. Rational Root Theorem</h4>
<ul style="color: rgb(255,255,255);">
<li>The Rational Root Theorem states that if a polynomial has integer coefficients, then any rational root, $rac{p}{q}$, must have 'p' as a factor of the constant term and 'q' as a factor of the leading coefficient.</li>
<li>In $x^3+2x^2-5x-6$, the constant term is -6 and the leading coefficient is 1.</li>
<li>Factors of -6 are: $\pm 1, \pm 2, \pm 3, \pm 6$.</li>
<li>Factors of 1 are: $\pm 1$.</li>
<li>Possible rational roots are: $\pm 1, \pm 2, \pm 3, \pm 6$.</li>
</ul>

<h4 style="color: rgb(255,255,255);">2. Testing Potential Roots</h4>
<ul style="color: rgb(255,255,255);">
<li>We substitute these possible roots into the polynomial to find which ones make $P(x) = 0$.</li>
<li>Test $x = -1$: $P(-1) = (-1)^3 + 2(-1)^2 - 5(-1) - 6 = -1 + 2 + 5 - 6 = 0$. Thus, $(x+1)$ is a factor.</li>
<li>Test $x = 2$: $P(2) = (2)^3 + 2(2)^2 - 5(2) - 6 = 8 + 8 - 10 - 6 = 0$. Thus, $(x-2)$ is a factor.</li>
<li>Test $x = -3$: $P(-3) = (-3)^3 + 2(-3)^2 - 5(-3) - 6 = -27 + 18 + 15 - 6 = 0$. Thus, $(x+3)$ is a factor.</li>
</ul>

<h4 style="color: rgb(255,255,255);">3. Polynomial Division or Synthetic Division</h4>
<ul style="color: rgb(255,255,255);">
<li>Once a root (say $x = -1$) is found, we know $(x+1)$ is a factor. We can then divide the original polynomial by $(x+1)$ to obtain a quadratic polynomial.</li>
<li>Using synthetic division with root -1:</li>
<pre>   -1 | 1   2   -5   -6
      |    -1   -1    6
      -----------------
        1   1   -6    0</pre>
<li>The resulting quadratic is $x^2 + x - 6$.</li>
<li>Factor the quadratic: $x^2 + x - 6 = (x+3)(x-2)$.</li>
<li>Combining the factors, we get $(x+1)(x+3)(x-2)$.</li>
</ul>

<h4 style="color: rgb(255,255,255);">4. Alternative Factoring (Grouping)</h4>
<ul style="color: rgb(255,255,255);">
<li>Sometimes, direct grouping can be applied if the terms align properly, though this is less common for general cubic polynomials.</li>
</ul>

<h4 style="color: rgb(255,255,255);">Key Concepts for Exams</h4>
<ul style="color: rgb(255,255,255);">
<li>Understanding the relationship between roots and factors of a polynomial.</li>
<li>Proficiency in applying the Rational Root Theorem.</li>
<li>Skill in performing polynomial division or synthetic division accurately.</li>
<li>Ability to factor quadratic expressions.</li>
</ul>

divident = (divisor x quotient) + reminder 
divident - remainder = (divisor x quotient )

ans = x-1

p(x)=ax²+bx+c
a(0)²+b(0)+c=4
c=4
a(1)²+b(1)+4=9
a(-1)²+b(-1)+4=5
a+b=5
a-b=1
2a=6 ; a=3
b=5-3 = 2

ans: 3x² + 2x + 4

x² - (ɑ+β) x + ɑβ
ɑ+β=1
ɑβ = -12
x²-x-12

(x-1) ia factor of p(x), if p(a)=0
p(1)=0
k(1)²-3(1)+k=0
2k-3 = 0
k=3/2

(x-a) is a factor of p(x) if p(a)=0
p(-2) = 3(-2)³-4(-2)²-6(-2)+7 = 21
p(2) = 3(2)³-4(2)²-6(2)+7 = 3
p(-1) = 3(-1)³-4(-1)²-6(-1)+7 = 6
p(1) = 3(1)³-4(1)²-6(1)+7 = 0
therefore ans (x-1)

x²-1 = (x+1)(x-1)
p(1)=1⁴⁷=1 = 1+1 = 2
p(-1)=(-1)⁴⁷+1=-1+1=0
let reminder = Ax + B
p(x) = (x+1)(x-1)q(x)+Ax+B
2=A+B
0=-A+B
2=2B ; B=1; A=1
reminder = Ax + B = x+ 1

p(x) ÷ (x-a) , then the reminder is p(a).
2x-1 = 0 ; 2x = 1 ; x= 1/2
reminder = p(1/2) 
=4(1/2)³ + 2(1/2) + 1/2 + 1/2 
=1/2 + 1/2+ 1/2+ 1/2 = 4/ 2 = 2

Let p(x) = x³-ax²+6x-a
reminder p(a) = a³ - a(a²) + 6a -a 
=a³ - a³ + 6a -a 
= 5a

let p(x)= x³ - ax² + 6x - a
reminder = p(a) = a³ - a(a)² + 6a  -a = a³ - a³ +5a = 5a

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