+0

# polynomial

0
484
6

p(x)=2x^4+3x^3-17x^2-27x-9, expresssed in factored form is p(x)=(x+1)(x-3)(ax+b)(x+c). what are the values of a, b and c? I need the work shown so i can apply it to other questions thanks :)

Guest Jun 24, 2014

#5
+8

(2x + 1)(x + 3) = 2(x + 1/2)(x + 3) = (2x + 6)(x + 1/2).

Guest Jun 24, 2014
#1
+93650
+8

I am sure Heureka ( or someone else) will come along with a beautifully laid out explanation but in the mean time

$$(x+1)(x-3)=x^2-2x-3$$

so you need to divide your polynomial by that to get the product of the 2 brackets containing a,b,and c.

You can so this with a polynomial long division.  There are lots of you tube videos explaining thiase techniques.  I can find one if you need me to.   (I've included one below)

after that you just factorise what is left by ordinary techniques to get the value of a,b and c.

---------------------------------------

Long division

You could also use synthetic division but if you do you have to divide by x+1 first and repeat the process for x-3  You cannot do both at once with synthetic division

Melody  Jun 24, 2014
#2
+20024
+8

$$\begin{array}{lcl} 2x^4+3x^3-17x^2-27x-9&=&(\underline{x}+1)(\underline{x}-3)(\underline{ax}+b)(\underline{x}+c) \\ 2x^4&=&x*x*ax*x\\ 2x^4&=&ax^4 \quad \mbox{(Coefficient comparison 2=a)} \end{array} \\ \boxed{a=2}$$

$$\begin{array}{rcl} 2x^4+3x^3-17x^2-27x-9&=&(x+\underline{1})(x-\underline{3})(ax+\underline{b})(x+\underline{c}) \\ -9&=&1*(-3)*b*c\\ -9&=&-3bc \end{array} \\ \boxed{c=\frac{3}{b}}$$

$$\begin{array}{lcl} (x+1)(x-3)(ax+b)(x+c)&=& (x+1)(x-3)(2x+b)(x+\frac{3}{b}) \quad | \quad x+\frac{3}{b}=0 \quad | *b \quad bx+3=0\\ &=&(x+1)(x-3)(2x+b)(bx+3) \end{array}$$

$$\begin{array}{lcl} 2x^4+3x^3-17x^2-27x-9&=&(\underline{x}+1)(\underline{x}-3)(\underline{2x}+b)(\underline{bx}+3) \\ 2x^4&=&x*x*2x*bx\\ 2x^4&=&2bx^4 \quad \mbox{(Coefficient comparison 2=2b)} \end{array} \\ \boxed{b=1}$$

$$\boxed{c=\frac{3}{b}=\frac{3}{1}=3}$$

$$\boxed{(x+1)(x-3)(ax+b)(x+c)=(x+1)(x-3)(2x+1)(x+3)}$$

heureka  Jun 24, 2014
#3
+27040
+8

Seems to me there are two sets of solutions here, unless I've screwed up somewhere (always a distinct possibility!). See below:

Alan  Jun 24, 2014
#4
+20024
+5

$$\begin{array}{rcccccccl} ( 2x^4 &+& 3x^3 &-& 17x^2 &-& 27x &-& 9 ) : (x^2-2x-3) = 2x^2 +7x+3\\ -(2x^4&-&4x^3&-&6x^2)&&&&\\ --&-&-&-&-&&&&\\ &&7x^3&-&11x^2&-&27x&&\\ &-(&7x^3&-&14x^2&-&21x)&&\\ &&-&-&-&-&-&&\\ &&&&3x^2&-&6x&-&9\\ &&&&-(3x^2&-&6x&-&9)\\ &&&&&&&&0\\ \end{array}$$

$$\begin{array}{rcl} 2x^4 + 3x^3 -17x^2 -27x -9 &=& (x+1)(x-3) (ax+b)(x+c) \\ &=&(x+1)(x-3)(2x^2+7x+3) \end{array}$$

$$\begin{array}{rcccrcc} (ax+b)(x+c)&=&2x^2&+&7x&+&3 \\ (ax+b)(x+c)&=&ax^2&+&(ca+b)x&+&bc \\ \end{array} \\\\ \mbox{Coefficient comparison:} \\ a=2\\ ca+b=7\\ bc=3$$

$$\\2c+b=7 \qquad \Rightarrow \quad b=7-2c\\ bc=3\qquad \Rightarrow \quad (7-2c)c=3\qquad \Rightarrow \quad 2c^2-7c+3=0$$

$$c_{1,2}=\frac{7\pm\sqrt{49-4*2*3}}{2*2}=\frac{7\pm\sqrt{25}}{4}=\frac{7\pm5}{4}\\ c=c_1=\frac{12}{4}=3\\ c=c_2= \frac{2}{4}=\frac{1}{2}\\ b=b_1=7-2c_1=7-2*3=1\\ b=b_2=7-2c_2=7-2*\frac{1}{2}=6$$

1.

$$\\2x^4+3x^3-17x^2-27x-9=(x+1)(x-3)(2x+b_1)(x+c_1)=(x+1)(x-3)(2x+1)(x+3)$$

2.

$$\\2x^4+3x^3-17x^2-27x-9=(x+1)(x-3)(2x+b_2)(x+c_2)=(x+1)(x-3)(2x+6)(x+\frac{1}{2})$$

heureka  Jun 24, 2014
#5
+8

(2x + 1)(x + 3) = 2(x + 1/2)(x + 3) = (2x + 6)(x + 1/2).

Guest Jun 24, 2014
#6
+27040
+5

Anonymous posted: (2x + 1)(x + 3) = 2(x + 1/2)(x + 3) = (2x + 6)(x + 1/2).

Correct! This shows the two polynomials are the same (though note that the original question was to find the values of a, b and c - so there are two sets of answers to the original question).

Alan  Jun 24, 2014